You are designing a lathe motor, part of which consists of a uniform cylinder wh

Question

You are designing a lathe motor, part of which consists of a uniform cylinder whose mass is 130 kg and whose radius is 0.60 m that is mounted so that it turns without friction on its axis, which is fixed. The cylinder is driven by a belt that wraps around its perimeter and exerts a constant torque. At t = 0, the cylinder's angular velocity is zero. At t = 25 s, its angular speed is 450 rev/min. (a) What is the magnitude of the cylinder's angular momentum at t = 25 s? kg·m2/s (b) At what rate is the angular momentum increasing? kg·m2/s2 (c) What is the magnitude of the torque acting on the cylinder? N·m (d) What is the magnitude of the frictional force acting on the rim of the cylinder? N

Explanation / Answer

given :
For solid cylinder, moment of Inertia,
I =0.5*m*r^2
= 0.5*130*(0.6)^2
=23.4 Kgm^2
wi = 0 rad/s
wf = 450 rev/min
= 450*2*pi rad/ 60 s
=47.124 rad/s
t= 25 s
a)
Momentum = I*w
= 23.4 * 47.124
=1102.7 Kgm^2/s
b)
dw/dt = (wf-wi)/t
= (47.124-0)/25
=1.885 rad/s^2
momentum = I*w
d/dt (momentum) = I* dw/dt
= 23.4* 1.885
= 44.11 Kgm^2/s^2
c)
T = rate of change of momentum
= 44.11 Kgm^2/s^2
= 44.11 Nm
d)
f= T/r
= 44.11 / 0.6
= 73.5 N

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