4 points | Previous Answers Tipler6 10 P044 You are designing a lathe motor, par

Question

4 points | Previous Answers Tipler6 10 P044 You are designing a lathe motor, part of which consists of a uniform cylinder whose mass is 95 kg and whose radius is 0.80 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 25 s, its angular speed is 600 rev/min. My NotesAsk Your (a) what is the magnitude of the cylinder's angular momentum at t-23 57 1909 2 kg-m2s (b) At what rate is the angular momentum increasing? km2s (c) What is the magnitude of the torque acting on the cylinder? 7636N-m d) What is the magnitude of the frictional force acting on the rim of the cylinder The angular momentum of the cylinder changes because a net torque acts on it. You can find the angular momentum at the given time from its definition, and the magnitude of the net torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque. eRonk 3 -12 points Tipler6 10 P 051 soin My Notes Ask Your A lazy Susan consists of a heavy plastic disk mounted on a frictionless bearing resting on a vertical shaft through its center. The cylinder has a radius R 20 cm and mass M0.24 kg. A cockroach (mass m-0.015 kg) is on the lazy Susan, at a distance of 10 cm from the center. Both the cockroach and the lazy Susan are initially at rest. The cockroach then walks along a circular path concentric with the axis of the lazy Susan at a constant distance of 10 cm from the axis of the shaft. If the speed of the cockroach with respect to the lazy Susan is 0.01 m/s, what is the speed of the cockroach with respect to the room? eDook

Explanation / Answer

Q2.

mass=m=95 kg

radius=r=0.8 m

moment of inertia of the cylinder=0.5*mass*radius^2=30.4 kg.m^2

at t=0, cylinder’s angular velocity=w0=0

at t=25 seconds, angular velocity =w=600 rev/min=62.832 rad/s

part a:

magnitude of cylinder’s angular momentum=moment of inertia * angular speed

=1910.1 kg.m^2/s

part b:

rate of change of angular momentum=moment of inertia*angular acceleration

=30.4*(62.832-0)/25

=76.404 kg.m^2/s^2

part c:

torque=rate of change of angular momentum=76.404 N.m

part d:

torque=force*radius

==>76.404=force*0.8

==>force=95.505 N

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