Question

An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 300 N applied to its edge causes the wheel to have an angular acceleration of 0.972 rad/s2.(a)What is the moment of inertia of the wheel (in kg · m2)?(b)What is the mass (in kg) of the wheel?(c)The wheel starts from rest and the tangential force remains constant over a time period of 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?

Answer 1

a) I = 101.8518 kg*m^2

b) M = 1870.5564 kg

c) w = 4.374 rad/s

Step-by-step explanation:

a) We know that:

T = I*a

F*r = I*a

where T is the torque, F is the force, r is the radius of the cylinder, I is the moment of inertia of the disk and a is the angular aceleration.

So:

(300 N)(0.33m) = I(0.972 rad/s^2)

solving for I:

I = 101.8518 kg*m^2

b) Adittionally,

I =
`(1)/(2)MR^2`

Where M is the mass and R is the radius.

So:

101.8518 kg*m^2 =
`(1)/(2)MR^2`

101.8518 kg*m^2 =
`(1)/(2)M(0.33)^2`

solving for M, we get:

M = 1870.5564 kg

c) Using the next equation

w = at

where w is the angular velocity, a is the angular aceleration and t is the time. Replacing the values, we get:

w = (0.972 rad/s^2)(4.5s)

w = 4.374 rad/s