A pulley has a moment of inertia of 5 kg m2 and a radius of 1.1 m . A cord is wr

Question

A pulley has a moment of inertia of 5 kg m2
and a radius of 1.1 m . A cord is wrapped over
the pulley and attached to a hanging object
on either end. Assume the cord does not slip,
the axle is frictionless, and the two hanging
objects have masses of 1 kg and 7 kg .
Find the acceleration of each mass. The
acceleration of gravity is 9.81 m/s
2
.
Answer in units of m/s
2

Find the force in the cord supporting the
smaller mass.
Answer in units of N

Find the force in the cord supporting the
larger mass.
Answer in units of N

Explanation / Answer

T1,T2 is tension 7*9.81-T1 =7a T2-1*9.81 =a adding above we get T2-T1 =8a-6*9.81 = T1-T2 =58.86-8a ..... and (T1-T2)r =I*alpha alpha =a/r hence T1-T2 =5/1.1^2 =4.13 hence 58.86-8a =4.13 hence a =6.84 m/s^2 putting value of a and finding tension in larger mass = T1=20.79 and that of smaller T2 =16.65

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