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

Answer:

$\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18$

Explanation:

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

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

F: force applied

R: radius of the wheel

I: moment of inertia

$\alpha=\frac{(120N)(0.80m)}{26.88kgm^2}=3.57\frac{rad}{s^2}$

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

$\theta=\frac{2\pi (0.8m)}{5.0m}=\approx1.00rad$

$\omega=\sqrt{2(3.57\frac{rad}{s^2})(1.00rad)}=2.67\frac{rad}{s}$

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

$K_R=\frac{1}{2}I\omega^2=(26.88)(2.67)^2 J = 188.63J$

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:

$\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18$

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