Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q74-Q77)

 Q#74

A small disc is set rolling with a speed v on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part?

Answer:
Let the required height be h.
M.I. of the disc, I = ½mr²
The angular speed ⍵ = v/r
The rotational K.E. =½I⍵² = ½(½mr²)(v²/r²) = ¼mv²
Translational K.E. = ½mv²
Total K.E. on the horizontal part = ¼mv² + ½mv² = ¾mv²
This total K.E. will be converted to P.E. at maximum height h = mgh. Equating,
mgh = ¾mv²
h = ¾(v²/g)

Q#75
A sphere starts rolling down an incline of inclination 𝛉, find the speed of its center when it has covered a distance l.

Answer:
When the sphere covers a distance of l on the incline it descends through a height l.sinθ. Loss of its P.E. = mglsinθ

Let the speed of its center be v and the angular speed ⍵ at the given position.
Linear K.E. = ½mv²

Rotational K.E. = ½I⍵² = ½(2mr²/5)(v²/r²) = mv²/5
Total K.E. = ½mv² + mv²/5
= 7mv²/10
Equating we get,

7mv²/10 = mglsinθ
v² = 10glsinθ/7
v = √(10glsinθ/7)

Q#76

A hollow Sphere is released from the top of an inclined plane of inclination θ. (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding? (b) Find the kinetic energy of the ball as it moves down a length l on the incline if the friction coefficient is half the value calculated in part (a).

Answer:
Let the friction coefficient be µ. If the weight of the hollow sphere is m the normal force on it by the plane = mgcosθ and hence the frictional force F = µmgcosθ. The component of weight along the plane = mgsinθ.
The torque on the ball due to frictional force about the point of contact = µmgr.cosθ {where r is the radius of the ball}

M.I. of the ball = 2mr²/3

(a) If the ball does not slide the angular acceleration of the ball,

α = µmgrcosθ/(2mr²/3)
= 3µgcosθ/2r

With this angular acceleration the linear acceleration of the CoM (here the center of the sphere)
a = αr = (3µgcosθ/2r)r
a = 3µgcosθ/2

If the sphere does not slide then the following condition should be satisfied

mgsinθ – F = ma
mgsinθ – µmgcosθ = m3µg(cosθ/2)
sinθ = µcosθ + 3µcosθ/2 = 5µcosθ/2
5µcosθ/2 = sinθ
µ = 2tanθ/5

(b) Now the frictional coefficient µ' = µ/2 = tanθ/5
Hence the frictional force
F = µmgcosθ
= mg.cosθ(tanθ/5)
= mgsinθ/5

Torque on the ball
Τ = Fr = mgrsinθ/5
α = Fr/I = (mgrsinθ/5)/(2mr²/3)
α = 3gsinθ/10r

Now mgsinθ – F = ma
mgsinθ – mgsinθ/5 = ma
a = 4gsinθ/5

After moving distance l down the plane
v² = 0²+ 2al
v = √(2al)

Also v = 0 + at
t = v/a = √(2al)/a = √(2l/a) = √(2l/4gsinθ/5)
= √(5l/2gsinθ)

And ⍵ = 0 + αt
= (3gsinθ/10r)√(5l/2gsinθ)
= 3{(5l)(gsinθ)}^1/2/10r√2

The kinetic energy of the ball after it moves down a length l on the plane
= ½mv² + ½I⍵²
= ½m(2al) + ½(2mr²/3)45l(gsinθ)/(200r²)
= ml(4gsinθ/5) + (45mlg)sinθ/600
= 4mlg(sinθ/5) + (3mlgsinθ/40)
= (35/40)mlgsinθ
= (7/8)mlgsinθ

Q#77
A Solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.

Answer:
Here our aim is to first find the velocity of the ball when it reaches the bottom of the cup. If the radius of the hemispherical cup is R, then K.E. of the ball at the bottom
= the loss in P.E. = mg(R – r)    {Where m is the mass of the ball and the CoM of the ball will be r above the bottom}

But it is also = ½mv² + ½I⍵²
= ½mv² + ½(2mr²/5)(v²/r²)
= ½mv² + mv²/5
= 7mv²/10

Equating the two we get,
7mv²/10 = mg(R – r)
v² = 10g(R – r)/7

Now the radius of curvature for CoM of the ball at the bottom = R – r
Hence the centrifugal force on the ball
P = mv²/(R – r)
= m{10g(R – r)/7}/(R – r}
= 10mg/7

Total force by the ball on the bottom of the cup = mg + P
= mg + 10mg/7
= 17mg/7
∴ Normal force by the cup on the ball when it reaches the bottom
N = 17mg/7   

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