Q: Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Linear momentum of particle, p=pxi^+pyj^+pzk^Position vector of the particle, r=xi^+yj^+zk^Angular momentum, l=r×p⟹lxi^+lyj^+lzk^=i^(ypz−zpy)−j^(xpz−zpx)+k^(xpy−ypx)Therefore on comparison of coefficients,lx=ypz−zpyly=zpx−xpzlz=xpy−ypxThe particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.z=pz=0Thus lx=0ly=0lz=xpy−ypxThus when particle is confined to move in the x-y plane, the angular momentum of particle is along the z-direction.

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 ?

Accepted Answer

In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.i) Sphere - Centre of Sphereii) Cylinder - Middle Point on axis of cylinderiii) Ring - At centre of ring (Outside the ring)iv) Cube - At point of intersection of diagnolsNo, 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˚(1A˚=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.

Accepted Answer

Let Mass of H atom =mMass of Cl atom =35.5mLet the centre of mass of the system lie at a distance x A˚ from the Cl atom.Distance of the centre of mass from the H atom =(1.27−x) A˚Let us assume that the centre of mass of the given molecule lies at the origin. Let Hydrogen lie to the left of the origin and chlorine to the right.Position of Hydrogen atom =−(1.27−x) A˚Position of chlorine atom =x A˚ Therefore, we can have:m+35.5m−m(1.27−x)+35.5mx​=0⟹−m(1.27−x)+35.5mx=0⟹x=35.5+11.27​=0.035 A˚Hence, the centre of mass of the HCl molecule lies 0.035 A˚ from the Cl atom.

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 ?

Accepted Answer

The child is running arbitrarily on a trolley moving with velocity V. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.

Question 4

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a×b.

Accepted Answer

Consider two vectors OK = vector ∣a∣and OM = vector∣b∣, inclined at an angle θ as shown in the following figure.In △OMN, we can write the relation:sinθ=OMMN​=∣∣​b∣∣​MN​⟹MN=∣∣​b∣∣​sinθ∣∣​a×b∣∣​=∣a∣∣∣​b∣∣​sinθ             =OK×MN=2×21​×OK×MN             =2× Area of △OMK⟹ Area of △ OMK=21​×∣∣​a×b∣∣​

Question 5

Show that a.(b×c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors, a, b and c.

Accepted Answer

A parallelopiped with origin O and sides a,b, and c is shown in the following figure.Volume of parallelopiped=abcLet n^ be a unit vector perpendicular to both b and c. Hence n^ and c have the same direction.Therefore, b×c=bcsinθ n^=bcn^a.(b×c)=a.(bcn^)                  =abc= Volume of parallelopiped

Question 6

Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px​, py​ and pz​. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Accepted Answer

Linear momentum of particle, p​=px​i^+py​j^​+pz​k^Position vector of the particle, r=xi^+yj^​+zk^Angular momentum, l=r×p​⟹lx​i^+ly​j^​+lz​k^=i^(ypz​−zpy​)−j^​(xpz​−zpx​)+k^(xpy​−ypx​)Therefore on comparison of coefficients,lx​=ypz​−zpy​ly​=zpx​−xpz​lz​=xpy​−ypx​The particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.z=pz​=0Thus lx​=0ly​=0lz​=xpy​−ypx​Thus when particle is confined to move in the x-y plane, the angular momentum of particle is along the z-direction.

Question 7

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

Accepted Answer

Let at a certain instant two particles be at points P and Q, as shown in the following figure.Consider a point R, which is at a distance y from point Q, i.e.,QR=y∴ PR=d−yAngular momentum of the system about point P:LP​=mv×0+mv×d=mvd......(i)Angular momentum of the system about point Q:LQ​=mv×d+mv×0=mvd....(ii)Angular momentum of the system about point R:LR​=mv×(d−y)+mv×y=mvd....(iii)Comparing equations (i), (ii), and (iii), we get:LP​=LQ​=LR​......(iv)We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

Question 8

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are 36.9o and 53.1o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.

Accepted Answer

The free body diagram of the bar is shown in the following figure.Length of the bar is ,l=2mT1​,T2​ be the tensions produced in the left and right strings respectively.At translational equilibrium, we have,T1​sin36.9∘=T2​sin53.1∘⟹T2​T1​​=34​For rotational equilibrium, on taking the torque about the centre of gravity, we have:T1​(cos36.9∘)×d=T2​cos53.1∘(2−d)Using both equations, d=0.72mHence, the centre of gravity of the given bar lies 0.72 m from its left end.

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. Detere the force exerted by the level ground on each front wheel and each back wheel.

Accepted Answer

Mass of car m=1800kgDistance between front and back axles d=1.8m Distance between center of gravity and front axle=1.05mLet Rb​ and Rf​ be the forces exerted by the level ground on the back and front wheels respectively.At translational equilibrium:Rf​+Rb​=mg=17640 N.....(i)For rotational equilibrium, on taking the torque about the C.G., we have:Rf​(1.05)=Rb​(1.8−1.05)⟹Rb​=1.4Rf​.......(ii)Solving (i) and (ii) givesRf​=7350 NRb​=10290 NThe force exerted on each front wheel =7350/2 = 3675 NThe force exerted on each back wheel =10290/2 = 5145N

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 2MR2/5, 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 be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Accepted Answer

(a) The moment of inertia (M.I.) of a sphere about its diameter=2MR2/5According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.The M.I. about a tangent of the sphere =2MR2/5+MR2=7MR2/5(b) The moment of inertia of a disc about its diameter =MR2/4According to the theorem of perpendicular axis, the moment of inertia of a planar body (laa) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.The M.I. of the disc about an axis normal to the disc through the center =MR2/4+MR2/4=MR2/2Now, Applying the theorem of parallel axes:The moment of inertia about an axis normal to the disc and passing through a point on its edge =MR2/2+MR2=3MR2/2

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?

Accepted Answer

The moment of inertia for the hollow cylinder=I1​=mr2The moment of inertia for the solid sphere=I2​=52​mr2For hollow sphere we have τ=I1​α1​For solid sphere we have τ=I2​α2​⟹α1​α2​​=I2​I1​​=25​>1Thus α2​>α1​ω(t)=ω0​+αtThe angular velocity(ω) at a certain time will be greater for solid sphere.

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?

Accepted Answer

The moment of inertia of a solid cylinder=mr2/2=21​×20×(0.25)2=0.625kgm2Therefore kinetic energy=21​Iω2=3125JAngular momentum ,L=Iω=0.625×100=62.5Js

Question 13

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/. 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. (b) Show that the childs new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Accepted Answer

(a) Since no external force acts on the boy, the angular momentum is a constant. Thus I1​ω1​=I2​ω2​ ⟹ω2​=I2​I1​​ω1​=25​×40rev/min=100 rev/min(b) Initial kinetic energy =E1​=21​I1​ω12​ Final kinetic energy =E2​=21​I2​ω22​  Thus E1​E2​​=I1​I2​​×ω12​ω22​​=2.5 The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

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.

Accepted Answer

The moment of inertia of the hollow cylinder about its geometric axis:I=mr2=3×0.42kgm2The Torque developed is given by:τ=rF=0.4×30Nm=12NmWe know that the relation between the torque and angular acceleration is:τ=Iα⟹α=25rad s−2Thus linear acceleration(a)=rα=0.4×25m/s2=10m/s2

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 N m. 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 frictional torque). Assume that the engine is 100% efficient.

Accepted Answer

Given: The angular speed of the rotor is 200 rad/s.The torque needed to be transmitted by the engine is 180 Nm.The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω, P=τω=180×200W=36kW

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.

Accepted Answer

Let mass per unit area of the original disc=σThus mass of original disc=M=σπR2Radius of smaller disc=R/2.Thus mass of the smaller disc=σπ(R/2)2=M/4After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:M(concentrated at O), and -M(=M/4) concentrated at O'(The negative sign indicates that this portion has been removed from the original disc.)Let x be the distance through which the centre of mass of the remaining portion shifts from point O.The relation between the centres of masses of two masses is given as:x=(m1​r1​+m2​r2​)/(m1​+m2​)=(M×0−(M/4)×(R/2))/(M−M/4)=−R/6(The negative sign indicates that the centre of mass gets shifted toward the left of point O)

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?

Accepted Answer

Let W and W′ be the respective weights of the metre stick and the coin.The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.Mass of the metre stick=m′Mass of each coin=5gWhen the coins are placed 12cm away from the end P, the centre of mass gets shifted by 5cm from point R toward the end P. The centre of mass is located at a distance of 45cm from point P.The net torque will be conserved for rotational equilibrium about point R.10g(45−12)−m′g(50−45)=0Thus m′=66g

Question 18

A solid sphere rolls down on 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?

Accepted Answer

(a) Total energy of sphere at the top= mghAt the bottom, the sphere has both translational and rotational energy=21​mv2+21​Iω2Using conservation of energy,mgh=21​mv2+21​Iω2......(i)For solid sphere I=52​mr2......(ii)Also we have the relation, v=rω.......(iii)Solving above equations gives v=710gh​​Clearly the speed of sphere at the bottom does not depend on the angle of inclination, both g and h are independent of the angle of inclination. (b) Assug rolling without slipping,Iα=Fr (7/5)mr2×a/r=mgsin(θ)ra=(5/7)gsin(θ)Since θ1​<θ2​,(c )Acceleration for the less inclined plane is less.Thus it takes longer time to reach the bottom along the plane with θ1​ 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?

Accepted Answer

Total energy of the hoop=Translational Kinetic energy+ Rotational Kinetic energy=21​mv2+21​Iω2For hoop, I=mr2Also v=rωThus total energy=21​mv2+21​mr2(rv​)2=mv2Hence the work required to stop the hoop=Its total energy=mv2=100×0.22J=4J

Question 20

The Oxygen Molecule Has a Mass of 5.30 × 10–26 Kg and a Moment of Inertia Of 1.94×10–46 Kg M2 About an Axis Through Its Centre Perpendicular to the Lines Joining the Two Atoms. Find the Average Angular Velocity of the Molecule.

Accepted Answer

Mass of an oxygen molecule, m=5.30×10−26 kg Moment of inertia, I=1.94×10−46 kg m2 Velocity of the oxygen molecule, v=500 m/s The separation between the two atoms of the oxygen molecule =2r Mass of each oxygen atom =2m​ Hence, moment of inertia I , is calculated as:​(2m​)r2+(2m​)r2=mr2r=mI​​5.36×10−261.94×10−46​​=0.60×10−10 m​ It is given that:​KErot ​=K Etrans " ′′21​Iω2=32​×21​mv2mr2ω2=32​mv2ω=32​​rv​=32​​×0.6×10−10500​=6.80×1012rad/s​

Chapter 7 : System of Particles and Rotational Motion