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

Answer 2

(a) The angular velocity is given as  $w_f= 0.04363 \frac{rad}{s^2}$

(b) The angular displacement $\theta =50.265 \ radians$

(c) The maximum tangential speed $V= 10.471\frac{m}{s}$

What will be the angular velocity, angular displacement, and max tangential speed?

It is given that

t = Time taken = 48 s

$w_f$ = Final angular velocity

$w-f=\dfrac{20\times2\pi}{60} =2.09439 \dfrac{rad}{s}$

$w_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

Now by using the equation of rotation

$w_f=w_i+\alpha t$

$\alpha=\dfrac{w_f-w_i}{t} = \dfrac{2.09439-0}{48} =0.043633 \frac{rad}{s^2}$

Now for angular displacement

$\theta =w_it+\dfrac{1}{2} \alpha t^2$

$\theta=0\timest+0.5\times0.043633\times48^2$

$\theta=50.265\ rad$

Maximum tangential speed is given by

$V=wr$

$V=2.0943\times5$

$V=10.4719 \frac{m}{s}$

Thus

(a) The angular velocity is given as  $w_f= 0.04363 \frac{rad}{s^2}$

(b) The angular displacement $\theta =50.265 \ radians$

(c) The maximum tangential speed $V= 10.471\frac{m}{s}$

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