Solutions to Exercises on Rotational Mechanics HC Verma's Concepts of Physics Part 1 (Q37-Q43)

 Q#37

The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0.20 kg-m². The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.

Answer:
Let the tension in the string on the plane be T and on the fixed side be T'.
T = ma = 1.0a = a         {where a is the acceleration of the block}
Acceleration of the pulley = a/2
Angular acceleration = a/2r = a/0.4 m/s²
The torque on the pulley = (T' – T)(0.20)

I = 0.20 kg-m²
α = T/I
a/0.4 = (T' – T)(0.20)/0.20 = T' – T
T – T' = –a/0.4 = –2.5a                                                (i)

Let the mass of the pulley be M

Mr²/2 = I

M = 2(0.20/0.2)0.2 = 10 kg

Now Mg
–T – T' = Ma/2
T + T' = Mg – Ma/2 = 10(9.8) – 10a/2 = 98 - 5a                   (ii)

Adding (i) and (ii)

2T = 98 – 5a – 2.5a
2a = 98 – 7.5a
9.5a = 98
a = 98/9.5 = 10.31 m/s² ≈ 10 m/s²

Q#38
The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.50 kg-m² about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

Answer:
Let the acceleration of the 4 kg block be a and the tension in the string T. If the tension in the  2 kg block be T' then we have,

4gsin45° – T = 4a
T = 4g/√2 – 4a

And T' – 2g.sin45° = 2a       {Same acceleration for 2 kg block}
T' = 2a + 2g/√2

Net torque on the pulley = (T – T')(0.10)

= 0.10(4g/√2 – 4a – 2a – 2g/√2)
= 0.10(2g/√2 – 6a)

Angular acceleration α = a/r = a/0.10
From Torque = Iα

0.10(2g/√2 – 6a) = 0.50(a/0.10) = 5a
2g/√2 – 6a = 50a

56a = 2g/√2 = 2(9.8)√2/2 = 9.8√2
a = 9.8(1.41/56) = 0.247 m/s² ≈ 0.25 m/s²

Q#39
Solve the previous problem if the friction coefficient between 2.0 kg block and the plane below it is 0.50 and the plane below the 4.0 kg block is frictionless.

Answer:
Given µ = 0.50
Hence the equation for 2 kg block is now

T' – 2gsin45° – µ(2gcos45°) = 2a
T' = 2a + 2g/√2 + µ(2g/√2)
T' = 2a + √2g + √2gµ

Now the torque is = Tr – T'r = (T – T')r

= (4g/√2 – 4a – 2a – √2g – √2gµ)0.10
= 0.10 (4 x 9.8/1.41 – 6a – 1.41 x 9.8 x 1.5)
= 0.10(27.80 – 20.73 – 6a) = 0.10(7.07 – 6a)

Angular acceleration α = a/r
Since Torque = Iα
0.10(7.07 – 6a) = 0.50(a/0.10) = 5a
0.707 – 0.60a = 5a
5.6a = 0.707
a = 0.707/5.6 = 0.126 ≈ 0.125 m/s²

Q#40
A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

Answer:
Let us draw a diagram as below

The weight of the stick may be taken as acting at the CoM of the rod, i.e. 50 cm from the left.

W = mg = 200g/1000 = 0.20g N.

The weight of the small object
W' = 20g/1000 = 0.02g N, acting at 70 cm i.e. 0.70 m from the left end.

Let tension in the left string = T and in the right string = T'
Since the stick is in balance, adding the forces in the vertical direction, we get

T + T' = W + W' = 0.20g + 0.02g = 0.22g                  (i)

Now the sum of the torques about end B,

T(1.0) – W(0.50) – W'(0.30) = 0    {Since the stick has no rotation}
T = 0.20g(0.50) + 0.02g(0.30)
T = 0.1g + 0.006g = 0.106g = 0.106(9.8) = 1.04 N

From (i),
T' = 0.22g – 1.04 = 0.22 x 9.8 – 1.04
T’ = 1.12 N

Q#41
A uniform ladder of length 10.0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work Safely?

Answer:
Let the normal force on the ladder at the lower end = N and at the upper end = N' as shown in the figure below.

AB = 10 m and AC = 8 m.
The force of friction on the lower end f_A = µN_A
The weight of ladder acts at the middle D = W = 16g N
The weight of the electrician P = 60g N
Since the rotation of the ladder is zero, hence the sum of torque about A = 0

N_B (10cos37°) – 16g(5 sin37°) – 60g (8sin37°) = 0
N_B = 411.6 N
                              
The sum of forces in the vertical direction,

N_A – 16g – 60g = 0
N_A = 76g = 76(9.8) = 744.8 N 

The sum of forces in the vertical direction,
f_A – N_B = 0
f_A = µN_A = N_B = 411.6 N
then
µ = N_B/N_A = 411.6 N/744.8 N

µ = 0.553

Q#42
Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.

Answer:
Let the maximum weight of the mechanic be Mg.
Equating the forces in the vertical direction

mg + Mg = N

Equating the sum of the torques about B

µN(10cos37°) – N(10sin37°) + 16g(5sin37°) + Mg(2sin37°) = 0
N{0.54(40/5) – 30/5} + 16(9.8)(5)(3/5) + M(9.8)(2)(3/5) = 0

–1.68N + 470.4 + 11.76 M = 0
–1.68(mg + Mg) + 470.4 + 11.76 M = 0
–1.68mg – 1.68Mg + 470.4 + 11.76M = 0
–4.70M + 470.4 – 263.42 = 0
M = 44.03 ≈ 44.0 kg

Q#43
A 6.5 m long ladder rest against a vertical wall reaching a height of 6.0 m. A 60 kg man stands halfway up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.

Answer:
Let us draw a diagram as below,

AB = 6.5 m and AC = 3.25 m.

The distance of the lower end of the ladder from the wall, AO

= √(6.5² - 6.0²) = 2.5 m

(a) The torque of the of the force exerted by the man on the ladder about the upper end
= 60g(2.5/2) = 75(9.81) = 740 Nm

(b) Let the Normal force by the ground be R and the frictional force = f_A. Normal reaction by the wall = N_B.

Equating forces in the vertical direction,

N_A = 60g = 60(9.8) = 588 N ≈ 590 N

Equating the torques about end A,

N_B(6) – 60g(2.5/2) = 0

N_B = 122.5 N

Equating forces in the horizontal direction,

f_A – N_B = 0
f_A = N_B = 122.5 N ≈ 120 N 

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