Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (49-53)

 Q#49

A small heavy block is attached to the lower end of a light rod of length l which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle.

Answer:
Let us take P.E. of the block at the lowest position be zero. If the velocity given be v, Then K.E. = ½mv²

So total energy at this level

= P.E + K.E = ½mv²

If the velocity at the highest position is v', K.E. = ½mv'² and P.E. = mg(2l) = 2mgl

Total energy = ½mv'² + 2mgl

Due to the conservation of energy, it will be same at both points. hence,

½mv² = ½mv'² + 2mgl
v² = v'² + 4gl
v = √(v'² + 4gl)

Since l and g are constants here, for v to be minimum v'² should be minimum, i.e. v' = 0. Putting it in above expression we get,
v = √(0 + 4gl) = √4gl = 2√(gl)

Q#50
Figure (8-E12) shows two blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth and light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g = 10 m/s².

Answer:  
Let the spring make an angle θ with the horizontal at the instance it breaks off the surface and the force due to spring T. If the velocity at this instant is v, then total K.E. of the system = 2(½mv²) =mv².
K.E. at initial position = 0
Change in K.E. = mv²

Given spring constant k = 40N/m and Natural length of the spring h = 40 cm = 0.40 m
Elongation of the spring x = hcosecθ – h
Hence T = kx = k(h.cosecθ – h)

Equating vertical forces on the block on the horizontal surface,

Tsinθ = mg → T = mg/sinθ
so, k(hcosecθ – h) = mg/sinθ

cosecθ – 1 = mg/khsinθ
1 – sinθ = mg/kh
Sinθ = 1 – mg/kh

= 1 – 0.32(10)/40(0.4O =  0.8

Elongation x = hcosecθ – h = 0.4/0.8 – 0.4 = 0.1 m

Displacement s = hcotθ = 0.4(3)/4 = 0.3 m (Since cosθ = √(1 – sin²θ) = √(1 – 0.8²) = √0.36 = 0.6  and cotθ = cosθ/sinθ = 0.6/0.8 = 3/4)

W.D. by all the forces on the system equals the change in K.E., so
–½kx² + mgs = mv²     (negative sign is due to the elongation opposite to the force)

v² = gs – ½kx²/m = 10(0.3) – ½(40)(0.1)²/0.32      (Given m = 320 g = 0.32 kg)

v² = 3 – 20(0.01)/0.32 = 2.37

v = 1.53 m/s

Q#51
One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (Figure 8-E13). Initially, the spring makes an angle of 37° with the vertical when the system is released from the rest. Find the speed of the ring when the spring becomes vertical.

Answer:
When the spring is at the given initial position its length = hsec37°
Elongation from natural length = hsec37° – h = 0.25h
P.E. stored in the spring = ½k(0.25h)²

When the spring comes in the vertical position, it retains its natural length and P.E. stored in it becomes zero but the ring which was at rest initially having its K.E. zero now has a kinetic energy due to its speed v at the final position. Since mechanical energy of the system will be conserved it means total P.E. of the spring is converted to K.E. of the ring =½mv²

Equating the two we get,

½mv² = ½k(0.25h)²
mv² = k(0.25h)²
v =0.25h√(k/m) = ¼h√(k/m)

Q#52
Figure (8-E14) shows a light rod of length l rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth h below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the block moves in a complete circle about the ring?

Answer:
Since the rod is rigidly attached to the block hence when it makes a complete circle its height above the ring at the highest point will be equal to l. If its speed at the highest point is v then its total energy here is equal to K.E.+P.E

= ½mv²+mgl

This total energy is due to the P.E. of the block at height h = mgh (taking the ring as the zero P.E. level).

Equating the two we have,

mgh = ½mv² + mgl
h = ½v²/g + l

Since l is fixed, for h to be minimum the first term ½v²/g should be minimum. It will be minimum when v = 0. Putting this value in the expression we have,

h = 0 + l
h = l

Q#53
The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity √(10gl) where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60° with the upward vertical.

Answer:
Let us consider the position of the bob when the string makes an angle θ with the upward horizontal. See figure below,

Let tension in the string be T, the mass of the bob = m, velocity = v.
Total centripetal force = T + mgcosθ
Since its velocity is v, centripetal acceleration = v²/l
Since Force = mass x acceleration

T + mgcosθ = mv²/l
T = m(v²/l – g cosθ) = m(v² – lgcosθ)/l                                  (i)

Let us take bob at rest as base for P.E. i.e. P.E. = 0
Velocity given at this point = √10gl

K.E. = ½m(10gl)
Total Energy = P.E.+K.E. = 5mgl                                              (ii)

Now K.E. at present level = ½mv²

P.E. at present level
= mg(l + lcosθ) = mgl(1 + cosθ)
Total Energy = ½mv² + mgl(1 + cosθ)                                    (iii)

Total energy will be conserved, hence equating (ii) and (iii) we get,
½mv² + mgl(1 + cosθ) = 5mgl
mv² + 2mgl(1 + cosθ) = 10mgl
mv² = 10mgl – 2mgl(1 + cosθ) = mgl(8 – 2cosθ)
v² = gl(8 – 2cosθ)

Putting this in (i)
T = m(v² – lgcosθ)/l = m{gl(8 – 2 cosθ) – lgcosθ)}/l
= m{8gl – 2glcosθ – glcosθ}/l
= m(8g – 3gcosθ)

(a) When the string is horizontal, θ = 90°
T = m(8g – 3gcos90°) = m[8g – 3g(0)] = 8mg

(b) When the bob is at its highest point. θ = 0°

T= m(8g – 3gcos0°) = m(8g – 3g) = 5mg

(c) When the string makes an angle 60° with the upward vertical, θ = 60°
T = m(8g – 3gcos60°) = m[8g – 3g(½)]
= m(8g – 1.5g) = 6.5mg

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