Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q82-Q86)

 Q#82

A thin spherical shell lying on a rough horizontal surface is hit by a cue in such a way that the line of action passes through the center of the shell. As a result, the shell starts moving with a linear speed v without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface.

Answer:
Let the linear speed be v' when it starts pure rolling, hence its angular speed about the point of contact, ⍵ = v'/r {r = radius and m = mass of the shell assumed}.
Initial angular momentum about the point of contact = mvr
Final angular momentum about the point of contact = I⍵

= (2mr²/3 + mr²)(v'/r)
= (5mr²/3)(v'/r)
= 5mrv'/3

Since the angular momentum will be conserved,
5mrv'/3 = mvr
v' = 3v/5

Q#83
A hollow sphere of radius R lies on a smooth horizontal surface. It is pulled by a horizontal force acting tangentially from the highest point. Find the distance traveled by the sphere during the time it makes one full rotation.

Answer:
Let the tangential force be T. The acceleration of the sphere, a = T/m, where m = mass of the sphere.
The torque of the force about the center of the sphere = TR.
The angular acceleration α = TR/I = 3TR/2mR² = 3T/2mR
For a full rotation θ = 2π
Ifthe time taken in full rotation = t,

θ = (0)t + ½αt²
½αt² = 2π
½(3T/2mR)t² = 2π
3Tt²/4mR = 2π
t² = 8πmR/3T

The distance travelled = ut + ½at²
= 0 + ½(T/m)(8πmR/3T)
= 4πR/3

Q#84
A solid sphere of mass 0.50 kg is kept on a horizontal surface. The coefficient of static friction between the surface in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer:
µ = 2/7, m = 0.50 kg. Let the force applied at the top = F.
The torque of the force about the point of the contact = F(2R) = 2FR
M.I. about the point of contact

= 2mR²/5 + mR² = 7mR²/5

The angular acceleration

α = 2FR/(7mR²/5)
= 10F/7mR

For the sphere not to slip the linear acceleration of the CoM, a = αR = 10F/7m
The force of friction = P = µmg {in the forward direction, because the limiting force F produces a torque that pushes the surface at contact in backward direction}

∴F + P = ma
F + µmg = m(10F/7m) = 10F/7
10F/7 – F = (2/7)mg
3F/7 = 2mg/7
F = 2mg/3 = 2(0.50)(9.8)/3 ≈ 3.3 N

Q#85
A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed v/R in the anticlockwise direction as shown in the figure (10-E16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

Answer:
(a) Let the mass of the sphere  = m.
If the linear speed of the sphere when rotation stops is V,

Then,

mvR – Iv/R = mVR
mvR – (2mR²/5)(v/R) = mVR
v – 2v/5 = V
V = (5v – 2v)/5
V = 3v/5

(b) When the pure rolling starts,
V' = ⍵'R
Now, mVR = I⍵' + mV'R = (2mR²/5)(V'/R) + mV'R
mVR = 2mV'R/5 + mV'R
V = 2V'/5 + V' = 7V'/5
7V'/5 = V = 3v/5
V' = (5/7)(3v/5) = 3v/7

Q#86
A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

Answer:
Since the collision is elastic, the rebound speed of the sphere = v. The vertical wall is smooth hence after rebound it will hold its angular momentum = I⍵ = Iv/ R.
The angular momentum just after the strike about the contact point = mvR – Iv/R
(-ve because both have opposite direction)

= mvR – (2mR²/5)(v/R)
= mvR – 2mvR/5
= 3mvR/5

If the speed of the sphere when pure rolling in backward direction starts = V, then the angular speed ⍵ = V/R.

Now the angular momentum = I⍵ + mVR

= (2mR²/5)(V/R) + mVR
= 2mVR/5 + mVR
= 7mVR/5

Since the angular momentum will be conserved, hence

7mVR/5 = 3mvR/5
7V = 3v
V = 3v/7 

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