Question

A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.350 kg .

A. Calculate its moment of inertia about its center.

B. Calculate the applied torque needed to accelerate it from rest to 1950 rpm in 4.00 s if it is known to slow down from 1750 rpm to rest in 57.5 s .

Answer 1

A.

Moment of inertia of cylinder about its center is given by,

I = 0.5*M*R^2

given, M = mass of cylinder = 0.350 kg

R = radius of cylinder = 8.50 cm = 0.085 m

then, I = 0.5*0.350*0.085^2

I = 1.26*10^-3 kg*m^2

B.

Now by torque balance,

_net = _applied + _friction = I*

then, _applied = I* - _friction

here, = dw/dt = (1950 rpm)/4.00 = (1950*2*pi/60)/4.0 = 51.05 rad/s^2

_friction = I*_friction = (1.26*10^-3)*(-1750*2*pi/60)/57.5 = -4.02*10^-3 N*m

then, _applied = (1.26*10^-3)*51.05 + 4.02*10^-3

_applied = 6.83*10^-2 N*m

"Let me know if you have any query."