Dog on a merry go round

the angular momentum of the whole system is conserved in the absence of external torque.Here the system(merry go round+dog)is not affected by an external torque.So the angular momentum of the system initially and finally is 0.

Let Wd be angular speed of dog wrt to ground and Wm be angular speed of merry go round wrt to ground. initially there is no rotation therefore the total angular momentum is 0. when the dog starts to go around ,the disk starts to go around in the opposite direction for the angular momentum to be conserved.

therefore Ld+Lm=0 Ld(angular momentum of dog)=Id Wd=M (R^2) Wd=1280 Wd Lm(angular momentum of disk)=Im Wm=0.5 M (R^2) Wd=16000*Wm

Ld+Lm=0

16000Wm=-1280*Wd Wd=-12.5Wm

let them meet after angle 'G'(angle turned by disk) after time t.

therefore (2pi-G)/Wd=G/Wm simplifying we get G=26.667 degrees

As the dog walks around the merry-go-round, angular momentum is conserved. Therefore, $I_d \omega_d + I_m \omega_m=0$ , where $I_d, I_m$ are the moments of inertia of the dog and merry-go-round and $\omega_d,\omega_m$ are the angular velocities. We can use this to solve for $\omega_d$ in terms of $\omega_m$ . The total relative angular velocity of the dog with respect to the merry-go-round is $\omega=\omega_d+\omega_m=(1+I_m/I_d)\omega_m$ . After t seconds the dog has gone around $2\pi$ radians, so we can solve for

$\omega_m=\frac{2\pi}{(1+\frac{I_m}{I_d})t}$ .

The distance the merry-go-round has traveled in that time is $\theta=\omega_m t$ , which yields $\theta=0.465~radians=26.67~degrees$ .

Let angular velocity of dog be $\omega_1$ and that of the ride is $\omega_2$ . By conservation of angular momentum, $1/2 \times 500 \times 8^2 \times \omega_2 = 20 \times 8^2 \times \omega_1$ . So, $2\omega_1=25\omega_2$ . Now relative angular velocity of dog with respect to ride = $\omega_1+\omega_2 =25/2 \times \omega_2+\omega_2 = 27/2 \times \omega_2$ . In t sec., dog moves X1 angle with respect to ride and ride moves X2 angle ( with respect to ground). Therefore $t = X1/(\omega_1+\omega_2) = X2/\omega_2$ . Putting X1 = 360 degrees, $\omega_1+\omega_2 = 27/2 \times \omega_2$ , we solve for X2=80/3 degrees =26.66 degrees.

Let $\omega_1, \omega_2$ be the angular velocities of the dog and the merry-go-round, respectively. By the conservation of angular momentum, $I_1 \omega_1=20\times8^2 \omega_1= \frac{1}{2} 500\times8^2 \omega_2 =I_2 \omega_2$ . Hence, $\omega_1=12.5\omega_2$ . Let $\theta_1$ be the angular displacement of the dog with respect to the purple horse and $\theta_2$ be the angular displacement of the dog wrt the ground. After some time t= $t=\frac{\theta_1}{\omega_1+\omega_2}=\frac{\theta_2}{\omega_2}$ , the dog meets the purple ride again, hence $\theta_1=360^ \circ$ . Solving the above equation for $\theta_2$ , we get $\theta_2=\frac{80}{3}=26.7^\circ$ , which is the desired value.

Let angular velocity of dog be (w1) and that of the ride is (w2). By conservation of angular momentum, 1/2 * 500 * 8^2 * (w2) = 20 * 8^2 * (w1) So, 2 (w1)=25 (w2) Now relative angular velocity of dog with respect to ride = (w1)+(w2) =25/2 * (w2)+(w2) = 27/2 * (w2) In t sec., dog moves (X1) angle with respect to ride and ride moves (X2) angle ( with respect to ground.) t = (X1)/((w1)+(w2)) = (X2)/(w2). Putting (X1) = 360 degrees, (w1)+(w2) = 27/2 * (w2), (X2)=80/3 degrees. Therefore the answer is 26.667 degrees.