Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (62-64)

 Q#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?

Answer:
(a) Height gained in reaching the Top

H = lsinθ + h = lsinθ + R – R cosθ

P.E. at the top = mgH = mg(lsinθ + R – R cosθ)

K.E. at top = 0

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

= mg(lsinθ + R – R cosθ)

Initial P.E. = 0
Initial K.E. = ½mv₀²
Initial total energy = ½mv₀²

Equating these two
½mv₀² = mg(l.sinθ + R – Rcosθ)

v₀² = 2g(lsinθ + R – R cosθ) = 2g{lsinθ + R(1 – cosθ)}

v₀ = √[2g{lsinθ + 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{lsinθ + R(1 – cosθ)} – 2g(lsinθ + R – R cosθ)]

= (m/R)6g{lsinθ + 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, gRcosθ = 2gR(1 – cosθ)

cosθ = 2 – 2 cosθ
3cosθ = 2

θ = cos⁻¹(2/3)

Q#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.

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(gRcosθ)
= (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 + cosθdθ/dt]
dv/dt = (R²g/l) [–cos(θ + l/R)dθ/dt + cosθdθ/dt]/v
dv/dt = (R²g/l)[ –cos(θ + l/R)dθ/dt + cosθdθ/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

Q#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.

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

Tangential force on the particle = macosθ + mgsinθ

So tangential acceleration = Force/mass = (macosθ + mgsinθ)/m

= acosθ + gsinθ, but it is also equal to dv/dt.

So, dv/dt = a cosθ + gsinθ

Multiplying both sides by v,

vdv/dt = acosθ v + gsinθv

vdv/dt = acosθ(Rdθ/dt) + gsinθ(Rdθ/dt)    {From (i)}

vdv = aRcosθdθ + gRsinθdθ

Integerating both sides,

v²/2 = aRsinθ – gRcosθ + 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 = aRsinθ – gRcosθ + gR
v² = 2R(asinθ – gcosθ + g)
v = [2R(asinθ – g cosθ + g)]^1/2   

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