Solutions to Problems on "Work and Energy" - H C Verma's Concepts of Physics, Part-I, Chapter-8, EXERCISES Q-55 TO Q-64

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WORK AND ENERGY:--
EXERCISES (55-64)

55. Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.

Figure for Q 55

Answer: Let the required compression of the spring be x.

P.E. stored in the spring = ½kx²

When the spring is released, the spring does a work on the block and changes its K.E. which is equal to the P.E. stored in the spring = ½kx². When the block reaches P, a part of this K.E. is converted to P.E. Let v be the velocity of the block at P.

It's energy at P =K.E.+P.E.

= ½mv²+mgR        (Its height above flat surface = R)

Hence, ½mv²+mgR = ½kx²

→mv²+2mgR =kx²    ------------------- (i)

Since the block presses the track with a force mg, that means Normal force on the block is also mg which due to its velocity should be mv²/R. Hence, mv²/R = mg

→mv² =mgR

Putting this value in (i), we get,

mgR+2mgR = kx²

→kx² = 3mgR

→x² = 3mgR/k

→x = √(3mgR)/k 

56. The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √(3gl). Find the angle rotated by the string before it comes slack. 

Answer: Assuming l = length of the string. Speed given to the bob = u = √(3gl)

K.E. given to the bob = ½mu² = ½m*3gl =3mgl/2 ---- (i)

The string will become slack at the point where the tension in the string T = 0. Suppose this point comes after rotating an angle θ and bob's speed here is v. If at this position the string makes an angle ⲫ with the upward vertical then ⲫ = 180°-θ. Its height from the original position A is = h. (See the figure below.)

Figure for Q 56

 

h = AC + DC = l + l.cosⲫ = l(1 + cosⲫ)

Taking the level of original position of the bob as zero P.E. level, P.E. of the bob at final position = mg(l+l.cosⲫ) 

=mgl(1+cosⲫ) 

And K.E. = ½mv²

Total mechanical energy at this point = ½mv²+mgl(1+cosⲫ)

Since total mechanical energy will be conserved, hence (i) and (ii) will be equal.

Equating (i) and (ii) we get,

½mv²+mgl(1+cosⲫ) = 3mgl/2

→v² + 2gl(1+cosⲫ) = 3gl        ------------- (iii)

Consider the forces at the final position. Force along the string towards center = T+mg.cosⲫ and centripetal acceleration mv²/l. So,

T+mg.cosⲫ = mv²/l

→mv²/l = mg.cosⲫ           (since here T=0)

→v² = gl.cosⲫ

Putting this value in (iii)

gl.cosⲫ + 2gl(1+cosⲫ) = 3gl

→gl.cosⲫ + 2gl.cosⲫ = 3gl - 2gl = gl

→3gl.cosⲫ = gl

→3.cosⲫ = 1

→cosⲫ = 1/3

→cos(180°-θ) = 1/3

→ -cosθ = 1/3

→ cosθ = - 1/3

→θ = cos⁻¹(-1/3)

57. A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g =10 m/s².

Answer: Let us draw the figure for this problem.

Figure for Q 57

(a) u=√57 m/s, K.E. = ½mu² = 57m/2 J

As in the problem 56, at the point of string being slack, 

mv²/l = mg.cosⲫ

→v² = lg.cosⲫ

Total energy at final point = K.E. + P.E.

= ½mv² + mgl(1+cosⲫ) 

=½mlg.cosⲫ + mgl(1+cosⲫ) = 57m/2

(Since final energy will be equal to the initial energy)

→ lg.cosⲫ + 2gl(1+cosⲫ) = 57

→ 3gl.cosⲫ = 57 - 2gl

→ 3*10*1.5.cosⲫ = 57 - 2*10*1.5 =27

→ 45.cosⲫ = 27

→ cosⲫ = 27/45 = 3/5 = 0.60 = cos53°

→ ⲫ = 53°

(b) We have v² = lg.cosⲫ = 1.5*10*3/5 = 9

→ v = 3.0 m/s

(c) Once the string becomes slack, the bob will move like a projectile with initial velocity = 3.0 m/s and angle of projection ⲫ= 53°. For this projectile motion maximum height above the point of projection d = (v.sinⲫ)²/2g.

Let the maximum height reached above the point of suspension = x.  

x = l.cosⲫ + d

= l.cosⲫ + (v.sinⲫ)²/2g 

=1.5*3/5 + (3*4/5)²/2*10

= 0.9 + 12*12/5*5*20 =0.9 + 0.288 = 1.188

→x ≈ 1.2 m/s

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°.

Figure for Q 58

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.

Figure 58 (a)

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.

Figure for 58b

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+L/2*cosθ)

= ½mv²+mgL/2*(1+cosθ)

Total energy will remain the same, hence

½mv²+mgL/2*(1+cosθ) = mgL  

½v²+gL/2*(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 = mg.cosθ

→v² = (gL/2)*cosθ

Putting it in (i)

(gL/2)*cosθ +gL*(1+cosθ) = 2gL 

→ ½cosθ +(1+cosθ) = 2

→ 3/2*cosθ = 1

→ cosθ = 2/3

So maximum height reached by the bob above the lowest point

=L/2 + L/2*cosθ

=L/2*(1+2/3)

=L/2*(5/3)

=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)

Figure for 58c

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

→5L-5x = 2L

→3L = 5x

→5x = 3L

→x/L = 3/5 = 0.60

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 = mg.cosθ-N.

If the velocity of the particle be v, then

mg.cosθ-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,

mg.cosθ = mv²/r

→v² = gr.cosθ

Figure for Q 59

Change in height of the particle = r-r.cosθ

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

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

So,  mg(r-r.cosθ)=½mv² =½mgr.cosθ

→1-cosθ=½cosθ

→(3/2)*cosθ=1

→cosθ=2/3

→θ=cos⁻¹(2/3)     

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=mg.cos30°
=mg.√3/2
=√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
mg.cos(θ+30°) = mv²/R 
→ g.cos(θ+30°) = v²/R
→v² = gR.cos(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)
gR.cos(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

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

→v² = gR

→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-R.cosθ)

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-R.cosθ)
→u² = gR/4 + 2gR-2gR.cosθ = 9gR/4-2gR.cosθ ------- (i)
At this point N = 0.
Centripetal force = mg.cosθ
Hence, mu²/R = mg.cosθ
u² = gR.cosθ
Putting it (i)
gR.cosθ = 9gR/4 - 2gR.cosθ
→3cosθ = 9/4
→cosθ = 3/4
→θ = cos⁻¹(3/4) 

   

62. Figure (8-E17) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. (a) Find the minimum projection-speed v₀ for which the particle reaches the top of the track. (b) Assuming that the projection-speed is 2v₀ and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than v₀, where will the block lose contact with the track?

Figure for Q 62

Answer: (a) Height gained in reaching the Top

H=l.sinθ+h =l.sinθ+R-R.cosθ

P.E. at the top = mgH = mg(l.sinθ+R-R.cosθ) 

K.E. at top = 0

Total energy at the top = P.E.+K+E.

= mg(l.sinθ+R-R.cosθ) 

Initial P.E. = 0

Initial K.E. = ½mv₀²
Initial total energy = ½mv₀²
Equating these two
½mv₀² = mg(l.sinθ+R-R.cosθ)

Figure for Q 62

v₀² = 2g(l.sinθ+R-R.cosθ) =2g{l.sinθ+R(1-cosθ)}

v₀ = √[2g{l.sinθ+R(1-cosθ)}]

(b) Initial speed = 2v₀

Let the final speed = v

Total energy at the top =Toatal energy at the initial point

½mv²+mgH = ½m(2v₀)² =2mv₀²

→v²+2gH = 4v₀²

→v² = 4v₀²-2gH

Centripetal acceleration = v²/R
Hence force on the particle = mv²/R =(m/R)*(4v₀²-2gH)
=(m/R)*[8g{l.sinθ+R(1-cosθ)}-2g*(l.sinθ+R-R.cosθ)]
=(m/R)*6g{l.sinθ+R(1-cosθ)}
=6mg(1-cosθ+l.sinθ/R)
(c) If the initial speed is slightly more than v₀, it means the block just starts from the rest at top. Let the radius through the point at which the block leaves the surface makes an angle θ with the vertical and its velocity there is u. Normal force here = 0. Hence, 
mg.cosθ = mu²/R
→u² = gR.cosθ
K.E. =½mu², Change in P.E. =mg(R-R.cosθ). Equating,
u² = 2gR(1-cosθ)
or, gR.cosθ = 2gR(1-cosθ)
→cosθ = 2-2.cosθ
→3.cosθ = 2
→cosθ = 2/3
→θ = cos⁻¹(2/3)

63. A chain of length l and mass m lies on the surface of a smooth sphere of radius R>l with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the center of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ. (c) Find the tangential acceleration dv/dt of the chain when the chain starts sliding down.

Answer:  (a) See the figure below.

Figure for Q 63(a)

 Let the angle subtended by the chain at center = α
Length of the chain = l
Hence α = l/R
Let us take a very small length of the chain (at angle θ from the vertical) that subtends angle dθ at the center.
Length of this small length =Rdθ
Mass of this small length = (m/l)*Rdθ
Height of this small length above the center = R.cosθ
P.E. of this small length = (m/l)*Rdθ*g*R.cosθ
= (mR²g/l)*cosθdθ 
Integrating this over an angle α we get P.E. of the chain.
=∫(mR²g/l)*cosθdθ = (mR²g/l)*∫cosθdθ = (mR²g/l)[sinθ]₀α
= (mR²g/l)sinα = (mR²g/l)sin(l/R)

(b) K.E. of the chain = Change in P.E. of the chain.
To know the P.E. of the chain after it slides down an angle θ, we shall put the limits of integration from θ to β. Where these are the angles made by the radii touching the ends of the chain. β = θ+α
So P.E. at this point = (mR²g/l)[sinθ]^βθ = (mR²g/l)[sinβ-sinθ]
= (mR²g/l)[sin(θ+α)-sinθ]
= (mR²g/l)[sin(θ+l/R)-sinθ]
Initial P.E. of the chain = (mR²g/l)sin(l/R)
Change in P.E. of the chain 
= (mR²g/l)sin(l/R) - (mR²g/l)[sin(θ+l/R)-sinθ]
= (mR²g/l)[sin(l/R) - sin(θ+l/R)+sinθ] =K.E. of the chain after it slid through an angle θ.
(c) As we derived the expression for K.E. of the chain at an instant when slid angle = θ, we have K.E. =½mv² (where v is speed at this instant). So we have,
½mv² = (mR²g/l)[sin(l/R) - sin(θ+l/R)+sinθ]  ----(i)
Differentiating it with respect to time t,
→ ½m(2vdv/dt) = (mR²g/l)[0-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]
→ vdv/dt = (R²g/l) [-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]
→ dv/dt = (R²g/l) [-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]/v
→dv/dt=(R²g/l)[-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]/Rdθ/dt
→dv/dt=(Rg/l)[cosθ-cos(θ+l/R)]
→dv/dt=(Rg/l)[cos0-cos(0+l/R)] (when chain starts sliding θ=0)
→dv/dt=(Rg/l)[1-cos(l/R) =Tangential acceleration

64. A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.

Answer: Since the sphere is moving with a constant acceleration a, it is a non-inertial frame for the particle. When analyzing the particle we should apply a pseudo force ma on the particle opposite to the direction of a. When at angle θ, forces on it are mg downward and pseudo force ma horizontal and normal force N radial outward as shown in the figure.

Figure for Q 64

Let the speed of the particle = v =Rdθ/dt  ------ (i)

Tangential force on the particle = ma.cosθ+mg.sinθ

So tangential acceleration = Force/mass = (ma.cosθ+mg.sinθ)/m

= a.cosθ+g.sinθ, but it is also equal to dv/dt.

So, dv/dt = a.cosθ+g.sinθ

 Multiplying both sides by v,

vdv/dt = a.cosθ.v+g.sinθ.v

→vdv/dt = a.cosθ(Rdθ/dt)+g.sinθ(Rdθ/dt)    {From (i)}

→vdv = aR.cosθdθ+gR.sinθdθ

Integerating both sides,

v²/2=aR.sinθ-gR.cosθ+C         ------------------(ii)

(where C is the integral constant)

To know C, we apply boundary condition. 

At θ = 0, v=0, because it starts from rest. Putting in (ii),

0 = 0-gR+C

→C=gR

So now expression (ii) is,

v²/2=aR.sinθ-gR.cosθ+gR

→v² =2R(a.sinθ-g.cosθ+g)

→v=[2R(a.sinθ-g.cosθ+g)]1/2

===<<<O>>>===

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HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY

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