A solid sphere of radius = 8 cm and mass = 5 kg is sitting on a level table and

Question

A solid sphere of radius = 8 cm and mass = 5 kg is sitting on a level table and is attached with a
string over a pulley (radius = 2 cm and mass = .8 kg) to a hanging mass (m = 3 kg) which is 4
meters above the ground.
a) Calculate the linear acceleration of the sphere and the falling block, (a) sphere, (a) block.
b) Calculate the angular acceleration of the pulley and the sphere, (alpha)pulley, (alpha)sphere

ans:a) asphere = 2.85 m/s2 ablock = 5.7 m/s2
b) (alpha)pulley = 285 rad/s2
(alpha)sphere = 35.63 rad/s2

Explanation / Answer

Let T1 be the tenison in the string attached to block and T2 be the tension in the string attached to the sphere and F be the friction force on the sphere.

For block: mbg - T1 = mbab....................(1)

For sphere: T2 - F = msas.......................(2)

Torque on sphere = F*Rs = Iss.....................(3)

Torque on pulley = (T1-T2)*Rp = Ipp......................(4)

We have inertia of sphere Is = 2/5*msRs2 and inertia of pulley Ip = 1/2*mpRp2

Further, ab will be equal to tangenital acceleration of pulley. Thus, ap = Rpp

And, as = Rss

Adding (1) and (2),

(T1-T2) = mb(g-ab) - F - msas

From (3), F = Isas/Rs2 = 2/5*msas

Now, using (4) we get, [mb(g-ab) - 2/5*msas - msas]*Rp = 1/2*mpapRp

mb(g-ab) - 7/5*msas = 1/2*mpap

3*(9.81-ab) - 7/5*5*as = 1/2*0.8*ap

7*as = 29.43 - 3*ab - 0.4*ap

Putting a = as = ab = ap,

a = 29.43/(7+3+0.4) = 2.83 m/s2

acceleration of block = 2a = 2*2.83 = 5.66 m/s2

p = a/Rp = 5.66/0.02 = 285 rad/s2

s = a/Rs = 2.83/0.08 = 35.63 rad/s2

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