Question

A wheel of radius R = 0.80 m is pulled by a rope looped around a frictionless axle. The total mass of the wheel and axle assembly is M = 70.0 kg and its moment of inertia is I = 26.88 kg.m^2. A constant horizontal force of magnitude F = 120 N moves the assembly through a horizontal distance d = 5.0 m starting from rest. The wheel rolls without slipping.What is the ratio of the wheel's rotational kinetic energy to its total kinetic energy at this instant?

Answer 1

`(K_T)/(K_R)=(600J)/(188.63J)=3.18`

Step-by-step explanation:

To find the rotational kinetic energy you first calculate the angular acceleration by using the following formula:

`\tau=I\alpha\\\\FR=I\alpha\\\\\alpha=(FR)/(I)`

F: force applied

R: radius of the wheel

I: moment of inertia

`\alpha=((120N)(0.80m))/(26.88kgm^2)=3.57(rad)/(s^2)`

With this value you calculate the angular velocity:

`\omega^2=\omega_o^2+2\alpha \theta\\`

you calculate how many radians the wheel run in 5.0m

`\theta=(2\pi (0.8m))/(5.0m)=\approx1.00rad`

`\omega=\sqrt{2(3.57(rad)/(s^2))(1.00rad)}=2.67(rad)/(s)`

Next, you use the formula for the rotational kinetic energy:

`K_R=(1)/(2)I\omega^2=(26.88)(2.67)^2 J = 188.63J`

For the transnational kinetic energy you use the following equation:

`W=\Delta K_T` (net work equals the change in the kinetic energy).

By replacing the you obtain:

`\Delta K_T=Fd=(120)(5.0)J=600J`

Finally, the ratio between translational rotational kinetic energy is:

`(K_T)/(K_R)=(600J)/(188.63J)=3.18`

hence, translational kinetic energy is three times the rotational kinetic energy.