If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify thesethe properties of the broom, but I think theythese are a reasonable interpretation of thefor a physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters). Their values don't really matter, they just change the constant in $ \ddot{\theta} = - c \cos \theta$, which only affects the time scale.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation with $cos$ has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters). Their values don't really matter, they just change the constant in $ \ddot{\theta} = - c \cos \theta$, which only affects the time scale.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify the properties of the broom, but I think these are a reasonable interpretation for a physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters). Their values don't really matter, they just change the constant in $ \ddot{\theta} = - c \cos \theta$, which only affects the time scale.
This formula, also derived on this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation with $cos$ has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters), though their. Their values don't affect the cutoff and angle behaviorreally matter, butthey just affectchange the axes onconstant in $ \ddot{\theta} = - c \cos \theta$, which only affects the graphstime scale.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters), though their values don't affect the cutoff and angle behavior, but just affect the axes on the graphs.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters). Their values don't really matter, they just change the constant in $ \ddot{\theta} = - c \cos \theta$, which only affects the time scale.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the rodbroom (I used 2 meters), though their values don't affect the cutoff and angle behavior, but just affect the axes on the graphs.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $L/2 \cos \theta$$\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the rodbroom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the rod (I used 2 meters), though their values don't affect the cutoff and angle behavior, but just affect the axes on the graphs.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $L/2 \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the rod and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.
If the broom's angle to the ground is less than 47.9°, it will hit the ground first. Otherwise, the key will.
This assumes that the end of the broom on the ground doesn't slide, making it act like a hinge, and also that the broom handle is a uniform rod. The puzzle doesn't specify these properties of the broom, but I think they are a reasonable interpretation of the physical model.
Below are plots of the height, velocity, and acceleration of the top end of the broom and the key for when the broom starts at 47.9°. They hit the ground at the same time. Note that the broom falls slower at first but catches up at the end to tie. If it had started lower, it would win, and if it started higher, it would lose.
Derivation
I got these plots and estimated the cutoff angle 47.9° by numerically solving the following differential equation in the angle $\theta$, where $\ddot{\theta}$ represents its second derivative in time, e.g. the angular acceleration: $$ \ddot{\theta} = - \frac{3 g}{2 L}\cos \theta$$ Here, $g$ is the gravitational acceleration ($9.8 \thinspace m/s^2$) and $L$ is the length of the broom (I used 2 meters), though their values don't affect the cutoff and angle behavior, but just affect the axes on the graphs.
This formula, also derived on this site this site, comes from considering the torque $\tau$ on the broom around its pivot on the ground via the gravitational force $mg$ acting on its center. This center is $\frac{L}{2} \cos \theta$ horizontal distance away, so $$\tau = \frac{mg L}{2} \cos \theta.$$ Then, we use that torque causes angular acceleration as $\tau = - I \ddot{\theta}$, where the moment of inertia for a uniform rod around it end is $ I = \frac{1}{3} M L^2$. This gives the formula above relating $\ddot{\theta}$ to $\cos \theta$. Since it's more intuitive to think about and plot height rather than angle, and we want to compare to the key, we relate the height $h$ of the top of the broom as $h = L \sin \theta$.
The site then solves for the angle $\theta$ where the initial downward acceleration $\ddot{h}$ matches the key's acceleration $-g$ due to gravity. This gives $\theta = 35.3^{\circ}$. But, this isn't the angle that makes them reach the ground at $h=0$ at the same time. The broom will accelerate faster during the smaller angles it passes through on its journey, so it will arrive first. Rather, we need to determine its whole trajectory $\theta(t)$ to find when it reaches zero.
Simulation
I doubt that such a second-order differential equation has an analytic solution, so I simulated it numerically. I used a time step of $0.2$ milliseconds, which I think is accurate enough. I compared it with a key starting from the corresponding initial height simulated the same way but under constant downward acceleration from gravity.
I searched for the initial angle $\theta(0)$ where the broom and key reach zero on the same simulation step, and found 47.9°. This might be a tiny bit off due to the discreteness of the simulation, but it should be very close.