Question

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her friends from rest to a frequency of 10 rpm in 9 seconds. Assume the merry-go-round is a uniform disc of radius 2.3 m and has a mass of 750 kg. Rachel (48 kg) and Tayler (46 kg) sit opposite each other on the edge of the ride.
(A) What for did Breanna push with?
(B) How much work did Breanna do?

Answer 1

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$