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 ยท 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?

Respuesta :

Answer:

a) I = 101.8518 kg*m^2

b) M = 1870.5564 kg

c) w = 4.374 rad/s

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 = [tex]\frac{1}{2}MR^2[/tex]

Where M is the mass and R is the radius.

So:

101.8518 kg*m^2 = [tex]\frac{1}{2}MR^2[/tex]

101.8518 kg*m^2 = [tex]\frac{1}{2}M(0.33)^2[/tex]

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