Chapter – 6
System of Particles And Rotational Motion
Question 1. Give the location of the centre of mass of a
(i) sphere,
(ii) cylinder,
(iii) ring, and
(iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
Answer:
In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass
Question 2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A (1 A = 10 – 10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Answer:
Let us choose the nucleus of the hydrogen atom as the origin for measuring distance. Mass of hydrogen atom,m1= 1 unit (say) Since cholorine atom is 35.5 times as massive as hydrogen atom,.•. mass of cholorine atom, m2 = 35.5 units
Question 3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Answer:
When the child gets up and runs about on the trolley, the speed of the centre of mass of the trolley and child remains unchanged irrespective of the manner of motion of child. It is because here child and trolley constitute one single system and forces involved are purely internal forces. As there is no external force, there is no change in momentum of the system and velocity remains unchanged.
Question 9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Answer:
Let F1 and F2 be the forces exerted by the level ground on front wheels and back wheels respectively.Considering rotational equilibrium about the front wheels,F2 x 1.8 = mg x 1.05 or F2 = 1.05/1.8 x 1800 x 9.8 N =10290 N Force on each back wheel is =10290/2 N or 5145 N.Considering rotational equilibrium about the back wheels.F1 x 1.8 = mg (1.8 – 1.05) = 0.75 x 1800 x 9.8or F1=0.75 x 1800 x 9.8/1.8 = 7350 NForce on each front wheel is 7350/2 N or 3675 N.
Question 10.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.
Answer:
(a) Moment of inertia of sphere about any diameter = 2/5 MR2Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2(i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
Question 11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
Answer:
Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are I1 = MR2 and I2 = 2/5 MR 2 Let τ be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations α1 and α2 respectively. Then τ=I1 α1 = I2 α2The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time.
Question 12. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Answer:
M = 20 kg Angular speed, w = 100 rad s-1; R = 0.25 m Moment of inertia of the cylinder about its axis=1/2 MR 2 = 1/2 x 20 (0.25)2 kg m2 = 0.625 kg m 2 Rotational kinetic energy,Er = 1/2 Iw 2 = 1/2 x 0.625 x (100)2 J = 3125 J Angular momentum,L = Iw = 0.625 x 100 Js= 62.5 Js.
Question 13.
(a) A child stands at the centre of a turntable with his arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction
Answer:
(a) Suppose, initial moment of inertia of the child is I1 Then final moment of inertia,
Question 14. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
Answer:
Here, M = 3 kg, R = 40 cm = 0.4 m Moment of inertia of the hollow cylinder about its axis.I = MR 2 = 3(0.4)2 = 0.48 kg m2
Question 15. To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?Note: Uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter fricitional torque). Assume that the engine is 100 efficient.
Answer:
Here, a = 200 rad s-1; Torque, τ= 180 N – m Since,Power, P = Torque (τ) x angular speed (w)= 180 x 200 = 36000 watt = 36 KW.
Question 16. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc.Locate the centre of gravity of the resulting flat body.
Answer: Let from a bigger uniform disc of radius R with centre O a smaller circular hole of radius R/2 with its centre at O1 (where R OO1 = R/2) is cut out. Let centre of gravity or the centre of mass of remaining flat body be at O2, where OO2 = x. If σ be mass per unit area, then mass of whole disc M1 = πR2σ and mass of cut out part .1i.e., O2 is at a distance R/6 from centre of disc on diametrically opposite side to centre of hole.
Question 17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Answer:
Let m be the mass of the stick concentrated at C, the 50 cm mark, see fig. For equilibrium about C, the 45 cm mark,10 g (45 – 12) = mg (50 – 45)10 g x 33 = mg x 5=> m = 10 x 33/5 or m = 66 grams.
Question 18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination,
(a) Will it reach the bottom with the same speed in each case?
(b) Will it take longer to roll down one plane than the other?
(c) If so, which one and why?
Answer:
(a) Using law of conservation of energy
(c) Clearly, the solid sphere will take longer to roll down the plane with smaller inclination.
Question 19. A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Answer:
Here, R = 2 m, M = 100 kgv = 20 cm/s = 0.2 m/s Total energy of the hoop =1/2Mv2 + 1/2Iw 2=1/2 Mv 2 + 1/2(MR2) w2=1/2 Mv 2 +1/2 Mv 2 =Mv 2 Work required to stop the hoop = total energy of the hoop W = Mv 2 = 100 (0.2)2= 4 Joule.
Question 21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Answer:
Here, θ= 30°, v = 5 m/ sLet the cylinder go up the plane up to a height h.From 1/2 mv 2 +1/2I W 2 = mgh
Question 22. As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied halfway up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be friction less and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m2 )(Hint: Consider the equilibrium of each side of the ladder separately.)
Answer:
The forces acting on the ladder are shown in Fig. 7.14.Here, IV = 40 kg = 40 x 9.8 N = 392 N, AB = AC = 1.6 m, BD = 1/2 x 1.6 m = 0.8 m,BF = 1.2 m and DE 0.5 m.1=(392 x (1-0.375))/1=245 N Now, it can be easily shown that tension in the string T = NB – NC = 245 – 147 = 98 N.
Question 23. A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minutes. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2.
(a) What is his new angular speed? (Neglect friction)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Answer:
Here, I1= 7.6 + 2 x 5 (0.9)2 = 15.7 kg m 2 w1 = 30 rp mI 2 = 7.6 + 2 x 5 (0.2)2 = 8.0 kg m 2w 2 = ?According to the principle of conservation of angular momentum, I 2 w 2= I 1 w 1 w 2= I 1/I2 w1= 15.7 x 30 /8.0 = 58.88 rpm No, kinetic energy is not conserved in the process. In fact, as moment of inertia decreases, K.E. of rotation increases. This change comes about as work is done by the man in bringing his arms closer to his body