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). a(sphere)=2.85m/s^2 a(block)=5.7m/s^2

b) Calculate the angular acceleration of the pulley and the sphere, alpha(pulley), alpha(sphere). alpha(pulley)=285rad/s^2. alpha(sphere)=35.63rad/s^2

c) Calculate the tensions in the upper and lower strings, T1 and T2. T1=10.00N T2=12.27 N

d) Calculate the static frictional force, fs. What is the minimum coefficient of static friction required for pure rolling motion? fs=4.28N coefficient of static frction=0.087

e) Calculate the Total work done on the sphere, pulley, block Wsphere, Wpulley, Wblock W(sphere)=40J, W(pulley)=9.08J, W(block)=68.52J

f) Calculate the final angular and linear speeds of the sphere, w(sphere), vfsphere. Wf(sphere)42.20rad/s, Vf(sphere)=3.38m/s

g) Calculate the final angular speed of the pulley, w(pulley). Wf(pulley)=338rad/s

h) Calculate the final linear speed of the falling block, vfblock. Vf(block)=6.76m/s

i) Check your answer by equating the initial potential energy to the final total kinetic energy PEi=117.6J KEf(total)=117.6J

I need help with the step-by-step.

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

The tensions in the upper and lower strings, T1 and T2 are
T1 = m1 * g
and T2 = (m1 + m2) * g

f =3g where g is gravity

m is mass of sphere

f-fs =ma ...1
also fs*r =I

for pure rolling r =a

fs*r =Ia/r

here for solid sphere I =(2/5)mr^2

fs*r =(2/5)mr^2 *a/r

fs =(2/5)ma

putting thi in equation 1

we get

f-(2/5)ma =ma

a= (5/7)f =15g/7 where g is gravity

friction force =fs =(2/5)m*(15g/7) =6mg/7 = 4.2 N

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