Question

A counterweight of mass m = 4.90 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 = 2.50 kg. The spokes have negligible mass.

(a) What is the net torque on the system about the axle of the pulley?

magnitude
direction	to the right along the axis of rotation to the left along the axis of rotation

(b) When the counterweight has a speed v, the pulley has an angular speed ω = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley.

(c) Using your result from (b) and = d/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.)
m/s²

Answer 1

1. a) mg-T-ma Tr= Ida I a I = Mr2 2:50 x(7)²x104 I = I a I s Hr?a I = 1 2.25x103 T= Mrexa Toma mg - Ma-ma mg = a a =) 4-9089.8 » 6.48 m/s2 4.90 +2.50 Him 16.22N T=Ma = 2.50x 6.48 cs Scanned ith canned Daith Ty > 16:22x7x1012a 1.13 Nm right qarus of rotation CamScanner

D Li = Iw - IV 뺨 풍a | - 이 는가 들 P*r = a. 키 Cusmi2 aㅋ 113 x Yx10 12.25x153_ Scanned with CamScanner