Question

A horizontal 803-N merry-go-round is a solid disk of radius 1.41 m and is started from rest by a constant horizontal force of 49.7 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 2.90 s.

Answer 1

Answer:

The kinetic energy of the disk is 254.4 J.

Explanation:

The kinetic energy of the disk is given by:

$K = \frac{1}{2}I \omega^{2}$

The moment of inertia of the solid disk is:

$I = \frac{1}{2}mr^{2}$

The mass is:

$m = \frac{P}{g}$

$I = \frac{Pr^{2}}{2g} = \frac{803 N*(1.41 m)^{2}}{2*9.81 m/s^{2}} = 81.4 kgm^{2}$

Now, we need to find the angular acceleration as follows:

$\tau = I \alpha$

Also, the torque is related to the tangential force:

$\tau = F\times r$

$F\times r = I \alpha$

$\alpha = \frac{F \times r}{I} = \frac{49.7 N*1.41 m}{81.4 kg*m^{2}} = 0.86 rad/s^{2}$

Now, we can find the angular speed:

$\omega_{f} = \omega_{i} + \alpha t$  

$\omega_{i} = 0$ since it is started from rest  

$\omega_{f} = 0.86 rad\s^{2}*2.90 s = 2.50 rad/s$

Finally, the kinetic energy of the disk is:

$K = \frac{1}{2}I \omega^{2} = \frac{1}{2}81.4 kgm^{2}*(2.50 rad/s)^{2} = 254.4 J$

Therefore, the kinetic energy of the disk is 254.4 J.

I hope it helps you!