Rotational Motion
NCERT Solution
Part 3
Question 22: As shown in this figure, the two sides of a step ladder BA and CA are 1.6 cm long and hinged at A. The rope DE 0.5 m is tied half way 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 frictionless 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 m/s2).
Answer: Following figure shows various forces:
NB and NC are the forces exerted on ladder by the floor respectively at points B and C and T is the tension in rope.
BA = CA = 1.6 m (given), DE = 0.5 m, BF = 1.2 m and mass of weight m = 40 kg
Let us draw a perpendicular from A on the floor BC so that it intersects DE at mid-point H.
Δ ABI and &Δ AIC are similar triangles
So, BI = IC
So, I is the mid-point of BC
DE || BC
BC = 2 × DE = 1 m
AF = BA – BF = 0.4 m ……..(1)
D is the mid-point of AB
So, we can write
AD `=1/2` × BA = 0.8 m……..(2)
From equations (1) and (2) we get
FE = 0.4 m
So, F is the mid-point of AD.
Since FG ||DH and F is the mid-point of AD
Hence, G is the mid-point of AH.
Δ AFG ∼ Δ ADH
So, `(FG)/(DH)=(AF)/(AD)`
Or, `(FG)/(DH)=(0.4)/(0.8)=1/2`
Or, FG = `1/2DH`
`=1/2xx0.25=0.125` m
In Δ ADH
AH = AD2 - DH2
`= 0.8^2 – 0.25^2 = (0.8 + 0.25)(0.8 – 0.25) = 1.05 × 0.55 = 0.5775`
Or, AH = `sqrt(0.5775)=0.76` m
For translational equilibrium of the ladder, upward forces should be equal to the downward forces.
So, NC + NB= mg = 392 ………….(3)
For rotational equilibrium of ladder, net moment about A is
`-N_B×Bl+mg×FG+N_C×Cl+T×AG-T×AG=0`
Or, `-N_B×0.5+40×9.8×0.125+N_C×0.5=0`
Or, `(N_C-N_B)xx0.5=49`
Or, `N_C-N_B=9.8` ……(4)
Adding equations (3) and (4) we get:
NC = 245 N and NB = 147 N
For rotational equilibrium of side AB, let us consider the moment about A
`-N_B×Bl+mg×FG+T×AG=0`
Or, `-245xx0.5+40xx9.8xx0.125+T×0.76=0`
Or, T = 96.7 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 minute. 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)
Answer: Moment of inertia of the system = 7.6 kg m2
When the man stretches his arms to 90 cm, moment of inertia
`=2mr^2`
`= 2 × 5 × 0.9^2 = 8.1` kg m2
So, initial moment of inertia of the system = li = 7.6 + 8.1 = 15.7 kg m2
Angular speed = 30 rev/min
Angular momentum `L_i=l_iω_i`
= 15.7 × 30 = 471 …………(1)
When the man folds his hands to a distance of 20 cm, the moment of inertia:
`=2mr^2`
= 2 × 5 × 0.22 = 0.4 kg m2
Final moment of inertia = 7.6 + 0.4 = 8 kg m2
Final angular speed = ωf
Final angular momentum Lf = lfωf = 8 ωf …………(2)
We know `l_iω_i=l_fω_f`
Or, `ω_f=(l_iω_f)/(l_f)`
`=(471)/8=58.875` rev/min
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Answer: In this process, kinetic energy is not conserved rather it increases with decrease in moment of inertia. For this, the additional kinetic energy comes from the work done by the man when he brings his hands towards himself.
Question 24: A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.
Answer: Given, mass of bullet m = 10 g = 0.01 kg, velocity v = 500 m/s, width of door L = 1 m, radius of door r = m/2, mass of door M = 12 kg
Angular momentum on the door α = mvr
`= 0.01 × 500 × ½=2.5` kg m2 s-1
Moment of inertia of the door `l=(ML^2)3`
`=1/3xx13xx1^2=4` kg m2
Now, `α = lω`
Or, `ω=(α)l`
`= (2.5)4=0.625` rad s-s
Question 25: Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc passing through the center), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident.
(a) What is the angular speed of the two-disc system?
Answer: Disc 1: moment of inertia = l1, angular speed = ω1
Disc 2: moment of inertia = l2, angular speed = ω2
Angular momentum of disc 1 = L1 = l1ω1
Angular momentum of disc 2 = L2 = l2ω1
Total angular momentum = L1 = l1ω1 + l2ω2
Moment of inertia of the two-disc system = l = l1 + l2
If ω is the angular speed of the system, then Li = LT
Or, `l_1ω_1+l_2ω_2=(l_1+l_2)ω`
So, `ω= (l_1ω_1+l_2ω_2)/(l_1+l_2)`
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for the loss in energy? Take ω1 ≠ ω2.
Answer: Kinetic energy of disc 1 `=E_1=1/2l_1ω_1^2`
Kinetic energy of disc 2 `=E_2=1/2l_2ω_2^2`
Total kinetic energy `=E=1/2(l_1ω_1^2+l_2ω_2^2)`
Moment of inertia of the system `=l=l_1+l_2`
Angular speed of the system = ω
Final kinetic energy `=E_f=1/2(l_1+l_2)ω^2`
`=1/2(l_1+l_2)(l_1ω_1+l_2ω_2)/(l_1+l_2^2)`
`=1/2(l_1ω_1+l_2ω_2)/((l_1+l_2))`
So, Ei - E1
`=1/2(l_1ω_1^2+l_2ω_2^2)-(1/2(l_1ω_1+l_2ω_2)^2)/(l_1+l_2)`
Solving this equation we get:
`=(l_1l_2(ω_1-ω_2)^2)/(2(l_1+l_2))`
As all the quantities on RHS positive
So, `E_i-E_f>0`
Or, `E_1>E_F`
The kinetic energy is reduced due to frictional force which comes into play when two discs come closer.
Question 26:
(a) Prove the theorem of perpendicular axes.
Answer: Theorem of perpendicular axes: The moment of inertia of a planar body 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.
Following figure shows a body with center O and a point mass m in the x-y plane.
Moment of inertia about x-axis `l_x=mx^2`, about y-axis `l_y=my^2` and about z-axis `l_z=m(x^2+y^2)^(1/2)`
`l_x+l_y=mx^2+my^2`
`=m(x^2+y^2)`
Or, `l_x+l_y=l_z` proved
`=m[(x^2+y^2)^(1/2)]^(1/2)`
(b) Prove the theorem of parallel axes.
Answer: 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 center of mass and the product of its mass and square of the distance between two parallel axes.
Let us assume that a rigid body is made up of n particles having masses m1, m2, m3 ………mn
Their perpendicular distances from the center of mass O are r1, r2, r3,………rn
Moment of inertia about axis RS passing through point O
`l_(RS)=Σm_i\r_i^2`
Since the perpendicular distance of mass mi from axis QP = a + ri hence moment of inertia about axis QP
`l_(QP)=Σm_i(a+r_i)^2`
`=Σm_i(a^2+r_i^2+2ar_i)`
`=Σm_1\a^2+Σm_ir_i^2+Σm+i2ar_i`
`=l+(RS)+Σm_ia^2+2Σm_i\ar_i^2`
Now, at the center of mass, the moment of all particles about the axis passing through the center of mass is zero.
2&Simga;m_i\ar_i=0`
Since a ≠ 0
Hence, `Σm_ir_i=0`
Also, Σ mi = M = Total mass of rigid body
So, `l_(QP)=l_(RS)Ma^2`
This proves the theorem.
Question 27: Prove the result that the velocity v of translation of a rolling body at the bottom of an inclined plane of a height h is given by:
`v^2=(2gh)/(1+(k^2)/(R^2))`
Using dynamical consideration (i.e. consideration of forces and torques).
Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Answer: Let us assume that a body (l = Mk2) is rolling down an inclined plane with initial velocity u and final velocity v.
We know: Loss in PE = Gain in translational KE + Gain in rotational KE
`Mgh=1/2mv^2+1/2lω^2`
`=1/2mv^2+1/2(mk^2)((v^2)/(R^2))`
Or, `Mgh=1/2mv^2(1+(k^2)/(R^2))`
Or, `v^2=(2gh)/(1+(k^2)/(R^2))`
Question 28: A disc rotating about its axis with angular speed ωO is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in this figure? Will the disc roll in the direction indicated?
Answer: Since `v=rω`
So, for point A: `v_A=Rω_0` (In the direction of arrow)
For point B: `v_B=Rω_0 (In opposite direction of arrow)
For point C: `v_C=R/2ω_0` (In the direction of arrow)
The disc will not roll down because friction is necessary for the same.
Question 29: Explain why friction is necessary to make the disc in this figure to roll down in the direction indicated.
Answer: We need a torque to roll a disc. The torque can be provided only by a tangential force. In this case, friction is the only tangential force present. So, friction is necessary to roll down the disc.
(a) Give the direction of frictional force at B, and at the sense of frictional torque, before perfect rolling begins?
Answer: Since velocity at point B is towards left so frictional force must be towards right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.
(b) What is the force of friction after perfect rolling begins.
Answer: As frictional force at B decreases the velocity of the point of contact B with the surface, perfect rolling begins only when velocity of point B becomes zero. The frictional force would become zero at this stage.
Question 30: A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic friction μk = 0.02.
Answer: Given, radii of ring and disc = r = 10 cm = 0.1 m, initial angular speed ω1 = 10 π rad s-1, coefficient of kinetic friction μk = 0.02 and initial velocity of both objects u = 0
Frictional force `F = ma`
Or, `μ_k\mg=ma`
Or, `a=μ_kg` …………(1)
Now, final velocity of objects can be calculated as follows:
`v=u+at=0+μ_k\g\t`
`=μ_k\g\t` …………(2)
Now, torque `τ=-lα`
Or, `μ_km\gr=-lα`
Or, `α=-(μ_km\gr)/l` …………….(3)
Angular speed can be obtained as follows:
`ω=ω_1+αt`
`=ω_1+((-μ_km\gr)/l)t` ……..(4)
Rolling starts when linear velocity `v=rω`
So, `v=r((ω_1-μ_1mg\rt)/l)`
Or, `μ_kg\t r((ω_1-μ_1mg\rt)/l)`
`=(rω_1-μ_k\mg\r^2t)/l` ………….(5)
For the ring:
`l=mr^2`
So, `μ_kg\t=(rω_1-μ_km\gr^2t)/(mr^2)`
`=rω_1-μ_kg\t`
Or, `2μ_kg\t=rω_1`
Or, `t=(rω_1)/(2μ_kg\t)`
`=(0.1xx10xx3.14)/(2xx0.2xx9.8=0.80)` s ………….(6)
For the disc:
`l=1/2mr^2`
So, `μ_kg\t=rω_1-(μ_km\gr^2t)/(1/2mr^2)`
`=rω_1-2μ_kg\t`
Or, `3μ_kg\t=rω_1`
Or, `t=(rω_1)/(3μ_kg)`
`=(0.1xx10.3.14)/(3xx0.2xx9.8)=0.53` s ……………….(7)
Question 31: A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μs = 0.25.
Answer: Given, mass of cylinder m = 10 kg, radius r = 15 cm = 0.15 m, coefficient of friction μk = 0.25, angle of inclination θ = 30°
Moment of inertia of a solid cylinder about its geometric axis
`l=1/2mr^2`
Following figure shows various forces acting on the cylinder.
Acceleration of cylinder can be calculated as follows:
`a=(mg si\nθ)/(m+l/(r^2))`
`=(mg si\nθ)/(m+(1/2mr^2)/(r^2))`
`=2/3g si\n 30°`
`=2/3xx9.8xx0.5=3.27` ms-2
(a) How much is the force of friction acting on the cylinder?
Answer: Net force can be calculated as follows: `f_(ne\t)=ma`
Or, mg sin 30° - f = ma
Or, f = mg sin 30° - ma
= 10 × 9.8 × 0.5 – 10 × 3.27
= 49 – 32.7 = 16.3 N
(b) What is the work done against friction during rolling?
Answer: Instantaneous point of contact with plane comes to rest during rolling. So, work done against frictional force is zero.
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?
Answer: This can be calculated as follows:
`μ=1/3` tan θ
Or, tan θ = 3μ = 3 × 0.25 = 0.75
So, θ = tan-1 (0.75) = 36.87°
Question 32: Read each statement below carefully, and state, with reasons, if it is true or false:
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
Answer: In case of rolling, the direction of motion of center of mass is backward. So, force of friction acts in forward direction because frictional force acts opposite to the direction of motion of the center of mass of the body. This statement is FALSE.
(b) The instantaneous speed of the point of contact during rolling is zero.
Answer: TRUE
(c) The instantaneous acceleration of the point of contact during rolling is zero.
Answer: This is FALSE statement because when a body is rolling, the instantaneous acceleration cannot be zero.
(d) For perfect rolling motion, work done against friction is zero.
Answer: One perfect rolling starts, force of friction becomes zero. So, work done against friction becomes zero. So, this is a TRUE statement.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer: Rolling can occur only in presence of friction. So, this statement is TRUE.