Question

A horizontal 791-N merry-go-round is a solid disk of radius 1.56 m and is started from rest by a constant horizontal force of 49.0 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 3.03 s.

Answer 1

Answer:

$K.E=273.5J$

Explanation:

Given data

$F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s$

To find

Kinetic Energy

Solution

The moment of inertia is given as:

$I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2$

The angular acceleration is given as:

$\alpha =\frac{T}{I}\\\alpha =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha =0.778rad/s^2$

Now the angular velocity is given by:

$w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s$

So the kinetic energy given as:

$K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J$