Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (58-61)

 Q#58

A simple pendulum of length L having a bob of mass m is deflected from its rest position by an angle θ and released (Figure 8-E16). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centered at the peg. (a) Assuming that initially, the bob has a height less than the peg, show that the maximum height reached by the bob equals its initial height. (b) If the pendulum is released with θ = 90° and x = L/2 find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a complete circle about the peg when the pendulum is released from θ = 90°.

Answer:
(a) Let the initial height of the bob above the rest position be h. Taking the rest position as zero P.E. level, Its initial P.E. = mgh.

Initial K.E. = 0 because speed is zero. So total Energy = mgh.

When the bob reaches the maximum height say h', its P.E. = mgh' and K.E. = 0 because again speed is zero. Now total energy = mgh'.

Since total energy will be conserved.

mgh' = mgh
h' = h

Hence proved.

(b) Taking the lowest position of the bob as zero potential energy level. At the point θ = 90°, P.E. of the bob = mgL, K.E. = 0, Hence total energy = mgL.

Let at the point where string becomes slack, its speed = v and the string makes θ with the upward vertical. Toatal energy at this point = K.E. + P.E.

= ½mv² + mg(L/2 + ½ Lcosθ)
= ½mv² + ½ mgL(1 + cosθ)

Total energy will remain the same, hence
½mv² + ½ mgL(1 + cosθ) = mgL
½v² + ½ gL(1 + cosθ) = gL
v² + gL(1 + cosθ) = 2gL                                               (i)

Here tension in the string T = 0 and only centripetal force = mg.cosθ.
Hence

mv²/½L = mgcosθ
v² = (gL/2)cosθ

Putting it in (i)

(gL/2)cosθ + gL(1 + cosθ) = 2gL
½cosθ + (1 + cosθ) = 2
1.5cosθ = 1
cosθ = 2/3

So maximum height reached by the bob above the lowest point

= L/2 + ½ Lcosθ
= ½ L(1 + 2/3)
= 5L/6

(c)   When the bob makes a complete circle in the above condition, the weight of the bob is the only centripetal force on the bob. Hence

mv²/(L – x) = mg
v² = g(L – x)

Equating the energies at top of the circle and at the initial point.

Mg{2(L – x)} + ½mv² = mgL
mg{2(L – x)} + ½mg(L – x) = mgL
2(L – x) + (L – x)/2 = L
5(L – x)/2 = L
5(L – x) = 2L
x/L = 3/5 = 0.60

Q#59
A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

Answer:
Let the radius to the particle make an angle θ with the upward vertical. If the normal force on the block is N and weight mg then centripetal force = mgcosθ – N
If the velocity of the particle be v, then

mgcosθ – N = mv²/r   (Where r is the radius of the sphere)
At the point where the particle leaves the surface, the normal force N = 0.

So, mgcosθ = mv²/r
v² = grcosθ

Change in height of the particle = r – r cosθ

Change in P.E. at the point it leaves the surface = mg(r – rcosθ)

This change in P.E. will result in increase of K.E. = ½mv²

So,  mg(r – r cosθ) = ½mv² = ½mgrcosθ

1 – cosθ = ½cosθ
(3/2) cosθ = 1
cosθ = 2/3
θ = cos⁻¹(2/3)    

Q#60
A particle of mass m is kept on a fixed smooth sphere of radius R at a position where the radius through the particle makes an angle 30° with the vertical. The particle is released from the position. (a) what is the force exerted by the sphere on the particle just after the release? (b) find the distance traveled by the particle before it leaves contact with the sphere.

Answer:
(a) Since the sphere is smooth, the force exerted by the sphere on the particle will be normal to the surface = N. Weight of the particle = mg (Downward)

Equating the force along the radius,

N = mgcos30° = √3mg/2

(b) Let it travel through an angle θ before it leaves the surface. At this point N = 0. If its velocity at this point is v, then

mgcos(θ + 30°) = mv²/R
gcos(θ + 30°) = v²/R
v² = gRcos(θ + 30°)                                                     (i)

Change in P.E. = mg{Rcos30° - Rcos(θ + 30°)}
Change in K.E. = ½mv²

Equating these two

½mv² = mg{Rcos30°- Rcos(θ + 30°)}
v² = 2g{Rcos30°- Rcos(θ + 30°)}

Putting the value of v² from (i)

gRcos(θ + 30°) = 2g{Rcos30°- Rcos(θ + 30°)}

cos(θ + 30°) = 2cos30° - 2cos(θ + 30°)

3cos(θ + 30°) = √3

cos(θ + 30°) = √3/3 = 1/√3 ≈ cos55°

θ = 55°- 30° = 25° = 0.43 radian

Hence distance traveled by the particle = Rθ = 0.43R

Q#61
A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of v for which the particle does not slip on the sphere? (c) Assuming the velocity v to be half the minimum calculated in part (b), find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Answer:
(a) At the top weight of the particle = mg (downward)
Normal force on the particle = N (Upward)
Net centripetal force = mg – N
Centripetal acceleration = mv²/R

Hence mg-N = mv²/R
N = mg – mv²/R

(b) For the limiting state when the particle is just to leave the surface N = 0

Hence 0 = mg – mv²/R
v²/R = g
v = √(gR)

(c) Given that v = ½√gR
K.E. = ½mv² = ½m(gR/4) = mgR/8

At the point of leaving the surface, the particle descends through R-R.cosθ. (Where θ is the angle by the radius through particle and upward vertical)

The decrease in P.E. = mg(R – Rcosθ)

If the speed of the particle at this point is u, its K.E. = ½mu²

Equating the energies at initial and final point,

½mu² = mgR/8 + mg(R – Rcosθ)
u² = gR/4 + 2gR – 2gR cosθ = 9gR/4 – 2gRcosθ                   (i)

At this point N = 0.

Centripetal force = mgcosθ
Hence, mu²/R = mgcosθ

u² = gRcosθ

Putting it (i)

gRcosθ = 9gR/4 – 2gRcosθ
3cosθ = 9/4
cosθ = 3/4
θ = cos⁻¹(3/4) 

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