PHYSICS UNIVERSITY

DAMON

2013-10-27 4:45 am

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, ). The pulley is a uniform disk with mass 10.7kg kg and radius 52.0cm cm and turns on frictionless bearings. You measure that the stone travels a distance 12.7m m during a time interval of 2.90s s starting from rest。

Q1 Find the mass of the stone.

Take the free fall acceleration to be 9.80m/s 2 9.80\;{\rm m/s^{2}}.

Q2
Find the tension in the wire.

Take the free fall acceleration to be 9.80m/s 2 9.80\;{\rm m/s^{2}}.

回答 (1)

天同

2013-10-27 7:41 am

✔ 最佳答案

Apply equation: s= ut + (1/2)at^2 to the stone
with s = 12.7 m, u = 0 m/s, t = 2.9 s, a =?
hence, 12.7 = (1/2)a.(2.9)^2
a = 3.02 m/s^2

Use: net force = mass x acceleration
mg - T = m.(3.02) --------------- (1)
where m is the mass of stone and T is the tension in the wire

Moment of inertia of pulley about its axis
= (1/2).(10.7).(0.52^2) kg.m^2 = 1.447 kg.m^2
Torque on pulley = 0.52T
Angular acceleration of pulley = 3.02/0.52 rad/s^2 = 5.808 rad/s^2
hence, 0.52T = (1.447).(5.808)
T = 16.16 N

Substitute T into (1),
mg - 16.16 = 3.02m
m = 16.16/(g - 3.02) kg = 2.38 kg

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