5) The flywheel of an engine has moment of inertia 2.75 kgm2 about its rotation

Question

5) The flywheel of an engine has moment of inertia 2.75 kgm2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 405 rev/min in 8.00 s, starting from rest?

6) A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point.

a) Find its angular acceleration due to this frictional force.

b) How long will it take to decrease its rotational speed by 24.5 rad/s?

Explanation / Answer

Convert that to rad/s --> 405 rev/min * .104 = 42.2 rad/ s
This is the final omega. The equation for rotational KE is (1/2) I * omega^2.

KE = (1/2) (2.75 kg * m^2) (42.2 rad/ s)^2 = 2193.2 J --> Round this to three sig digs:

Final answer: 2439J

The working formula is

Torque = Moment of inertia * angular acceleration

Torque = Frictional force * radius of sphere

Frictional force = 0.02 N (given)

Radius of sphere = 3/2 = 1.5 cm = 0.015 m

Therefore, Torque = 0.02 * 0.0175 = 0.00035 N-m

For a solid sphere of mass, the moment of inertia is

I = (2/5)mR^2

where

m = 225 gm (given) = 0.225 kg.
R = radius = 0.015 m

and substituting values,

I = (2/5)(0.225)(0.015)^2

I = 0.0000010

Solving for the angular acceleration,

Angular acceleration = torque/moment of inertia

Angular acceleration = 0.0003/0.0000101 = 29 rad/sec^2

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