Question

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 200p rad/s. Suppose that one such flywheel is a solid, uniform cylinder with a mass of 500 kg and a radius of 1.0 m.(a) What is the kinetic energy of the flywheel after charging

Answer 1

Answer:

The kinetic energy of the flywheel is 5000kJ.

Explanation:

We have that the moment of inertia of a solid and uniform cylinder is equal to:

$I=\frac{1}{2} mR^{2}$

Where m is the mass of the cylinder and R its radius.

Next, the rotational kinetic energy of an object rotating about an axis is given by:

$K_R=\frac{1}{2} I\omega^{2}$

Where ω is the angular velocity. From these two equations, we can derive that:

$K_R=\frac{1}{4} mR^{2} \omega^{2}$

Plugging the given values in this equation, we obtain:

$K_R=\frac{1}{4}(500kg)(1.0m)^{2}(200rad/s)^{2}=5*10^{6}J=5000kJ$

In words, the kinetic energy of the flywheel is 5000kJ.

Answer 2

Answer:

Explanation:

Radius, r = 1 m

Mass, m = 500 kg

w = 200 rad/s

Moment of inertia for a dolid symmetric cylinder, I = 1/2 × mr^2

KE = 1/2 × I × w^2

Therefore,

KE = 1/4 × m × w^2 × r ^2

= 1/4 × 500 × 200^2 × 1^2

= 5.0 × 10^6 J.