Question

A horizontal 800-N merry-go-round is a solid disk of radius 1.50 m and is started from rest by a constant horizontal force of 50.0 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 3.00 s.

Short answer

The kinetic energy of the disk after 3.00 s is,$\text{K}=276 \mathrm{J}$.

Step-by-step solution

Definition of energy consideration

According to the conservation of energy, the work done on the system = energy generated in the system.

The given values

$\begin{array}{c}W=800 \mathrm{N}\\ R=1.5 \mathrm{m}\\ {\omega }_i=0\\ F=50 \mathrm{N}\\ t=3 \mathrm{s}\end{array}$

Calculating the moment of inertia

The weight of the merry-go-round is found by:

$W=mg$

Solve for (m):

$\begin{array}{c}m=\frac{W}{g}\\ =\frac{800}{9.8}\\ =81.6 \text{kg}\end{array}$

The moment of inertia of the merry-go-round (solid disk):

$I=\frac{1}{2}M{R}^2$

Substituting the numerical value:

$\begin{array}{c}I=\frac{1}{2}(81.6)(1.5{)}^2\\ =91.8 \mathrm{kg}·{\mathrm{m}}^2\end{array}$

The merry-go-round as a rigid object under a net torque

$\begin{array}{rcl}\sum \tau & =& FR\\ & =& I\alpha \end{array}$

Solve for$\left(\alpha \right)$.

$\alpha =\frac{FR}{I}$

Substituting the numerical value

$\begin{array}{c}\alpha =\frac{\left(50\right)(1.5)}{91.8}\\ =0.817 \mathrm{rad}/{\mathrm{s}}^2\end{array}$

Model the merry-go-round as a rigid object under constant angular acceleration and use the following equation to find the final angular velocity

${\omega }_f={\omega }_i+\alpha t$

Substitute the numerical value

$\begin{array}{c}{\omega }_f=0+(0.817)\left(3\right)\\ =2.45 \mathrm{rad}/\mathrm{s}\end{array}$

The kinetic energy of the disk

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

Substitute the numerical value

$\begin{array}{c}K=\frac{1}{2}(91.8)(2.45{)}^2\\ =276 \mathrm{J}\end{array}$

Thus, the energy is $276 \mathrm{J}$.