AT&T; LTE 1:41 PM 33%. \" A counterweight of mass m-4.80 kg is attached to a lig

Question

AT&T; LTE 1:41 PM 33%. " A counterweight of mass m-4.80 kg is attached to a light cord that is wound around a pulley as shown in the figure below. The pulley is a thin hoop of radius R 7.00 cm and mass M-2.30 kg. The spokes have negligible mass. (a) What is the net torque on the system about the axle of the pulley? (b) When the counterweight has a speed v, the pulley has an angular speed ?·v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley. kg m)v (c) Using your result from (b) andd/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) Need Help? LaS A uniform solid disk of mass m-2.99 kg and radius r-0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 6.08 rad/s (a) Calculate the magnitude of the angular momentum of the disk when the axis of rotation passes through its center of mass. kg m2/s (b) What is the magnitude of the angular momentum when the axis of rotation passes through a point midway between the center and the rim? A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of mass 0.200 kg. The meterstick rotates on the surface of a frictionless,

Explanation / Answer

7)

a)

Torques on the axle

= 4.8 * 9.81 * 0.07 m-N

= 3.296 Nm ccw

b)

Angular momentum of the system

= angular momentum of the counterweight + angular momentum of the pulley

L = [4.8 * w * (0.07)^2] + [2.3 * (0.07)^2 w] = 0.03479 w kgm^2/s

c)

Torque = dL/dt

=> 3.296 = d/dt (0.03479 w)

=> dw/dt = 3.296/0.03479 rad/s^2

=> angular acceleration, dw/dt = 94.74 rad/s^2

=> acceleration of the counterweight

= dw/dt * R

= 94.74* 0.07 m/s^2

= 6.63 m/s^2.

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