Question

A child sits on a merry‑go‑round that has a diameter of 5.00 m. The child uses her legs to push the merry‑go‑round, making it go from rest to an angular speed of 20.0 rpm in a time of 48.0 s. What is the average angular acceleration ????avg of the merry‑go‑round in units of radians per second squared (rad/s2)? ????avg= ______rad/s2 What is the angular displacement Δtheta of the merry‑go‑round, in units of radians (rad), during the time the child pushes the merry‑go‑round? Δtheta= _______rad What is the maximum tangential speed ????max of the child if she rides on the edge of the platform? ????max=________m/s

Answer 1

Answer:

0.043633125 rad/s²

50.26536 radians

10.47195 m/s

Explanation:

t = Time taken = 48 s

$\omega_f$ = Final angular velocity

$\omega_f=20\times \frac{2\pi}{60}=2.09439$

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{2.09439-0}{48}\\\Rightarrow a=0.043633125\ rad/s^2$

The average angular acceleration is 0.043633125 rad/s²

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 0.043633125\times 48^2\\\Rightarrow s=50.26536\ rad$

The displacement of the merry‑go‑round is 50.26536 radians

Maximum tangential speed is given by

$v=\omega r\\\Rightarrow v=2.09439\times 5\\\Rightarrow v=10.47195\ m/s$

The maximum tangential speed of the child is 10.47195 m/s