A stone is suspended from the free end of a wire that is wrapped around the oute

Question

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, as shown in the figure (see the figure (Figure 1)). The pulley is a uniform disk with mass 11.5kg and radius 47.0Cm and turns on frictionless bearings. You measure that the stone travels a distance 12.5m during a time interval of 3.15s starting from rest. Part A Find the mass of the stone. Take the free fall acceleration to be 9.80 m/s^2. kg Part B Find the tension in the wire. Take the free fall acceleration to be 9.80 m/s^2. N

Explanation / Answer

from the kinematic equation

y = y0 + u + 1/2 a t^2

12.5 m = 0+0 + 1/2 a (3.15 s)^2

a = 2.519 m/s^2

the relation between torque and tenstion si

torque = TR = I alpha

TR = ( MR^2/2) (a/r)

T = Ma/2 = 11.5 kg ( 2.519 m/s^2)/2 = 14.48 N

(a)

The net force on the stone is

Fnet = mg- T

mg - ma = T

m = T/ ( g-a) = 14.48 N/(9.8 m/s^2 - 2.519 m/s^2) = 1.988 kg

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