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

`K.E=273.5J`

Step-by-step 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=((1)/(2) )MR^2\\I=(1)/(2)(F_(g)/g)R^2\\ I=(1)/(2)((791N)/(9.8m/s^2) )(1.56m)^2\\ I=98.21kg.m^2`

The angular acceleration is given as:

`\alpha =(T)/(I)\\\alpha =(F_(y)R)/(I)\\ \alpha =((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=((1)/(2) )Iw^2\\K.E=(1)/(2)(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J`