a block of mass m 1 = 12 kg hangs from a rope that goes over a pulley made of a

Question

a block of mass m 1 = 12 kg hangs from a rope that goes over a pulley made of a uniform disk Of mass mp =1.5 kg and is attached to a block of mass m2= 25 kg that sits on a horizontal surface with a coefficient of kinetic friction of .34 the hanging mass begins to fall, what is the magnitude of the acceleration of block 1 as it falls? a block of mass m 1 = 12 kg hangs from a rope that goes over a pulley made of a uniform disk Of mass mp =1.5 kg and is attached to a block of mass m2= 25 kg that sits on a horizontal surface with a coefficient of kinetic friction of .34 the hanging mass begins to fall, what is the magnitude of the acceleration of block 1 as it falls?

Explanation / Answer

he moment of inertia of the pulley. It is easy and straightforward to calculate so let us calculate it now.

I = (1/2) M R2

Apply Newton's Second Law to each of the three bodies:

First, for mass m1:

FNet, 1 = T1 - Ff1 = m1 a

Ff1 = FN1 = m1 g

T1 - m1 g = m1 a

Now look at the torques on the pulley. This is not the usual "lightweight pulley" we have encountered before. You might think of this as a flywheel. It has a moment of inertia that is large enough that it can not be neglected. This also means the tension in the rope on one side of it will be different than the tension on the other side. Hence, we have labeled the tensions at T1 and T2.

net = R T2 - R T1 = I

T2 - T1 = ( I / R)

Now we are ready for the forces on m2, the mass on the incline.

Fnet = w2 sin 30o - Ff2 - T2 = m2 a

Fnet = 0.866 w2 - Ff2 - T2 = m2 a

Ff2 = FN2 = w2 cos 30o = m2 g cos 30o

Recall that the linear acceleration of the blocks a is closely related to the angular acceleration of the pulley (or flywheel) ,

a = r

or

= a /r

Now we have three equations with three unknowns, T1, T2, and a.

T1 - m1 g = m1 a

T2 - T1 = ( I / R) = ( I / R ) ( a / R ) = ( I / R2 ) a

0.866 m2 g - 0.50 m2 g - T2 = m2 a

Now we can put in numbers and then solve for the unknowns.

T1 - (0.36) (2.0 kg) (9.8 m/s2) = (2.0 kg) a

T2 - T1 = [ 0.3125 kg m2 / (0.25 m)2 ] a = 5 kg a

0.866 (6.0 kg) (9.8 m/s2) - 0.50 (0.36) (6.0 kg) (9.8 m/s2) - T2 = (6.0 kg) a

50.92 N - 10.58 N - T 2 = (6.0 kg) a

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