Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Subjective Questions
Moment of inertia

Ques 1: Four thin rods each of mass m and length l are joined to make a square. Find moment of inertia of all the four rods about any side of the square.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
I = I1 + I2 + I3 + I4
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 2: A mass of 1 kg is placed at (1 m, 2 m, 0). Another mass of 2 kg is placed at (3 m, 4 m, 0). Find moment of inertia of both the masses about z-axis.
Ans: I = I1 + I2
= 1 × (12 + 22) + 2(32 + 42) = 55 kg-m

Ques 3: Moment of inertia of a uniform rod of mass m and length l is 7/12 ml2  about a line perpendicular to the rod.
Find the distance of this line from the middle point of the rod.
Ans: Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev  

Ques 4: Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 5: Radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Find its radius of gyration about a parallel axis through its centre of mass.
Ans: Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 6: Two point masses m1 and m2 are joined by a weightless rod of length r. Calculate the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the rod.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Substituting the values we get,
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 7: Linear mass density (mass/length) of a rod depends on the distance from one end (say λ) as λx = (αx + b). Here, α and β are constants. Find the moment of inertia of this rod about an axis passing through A and perpendicular to the rod. Length of the rod is l.
Ans:
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
dI = dmx2 = x2λdx = x2(αx + β)dx

(αx3 + βx2) dx
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Angular Velocity
Ques 8: Find angular speed of second's clock.
Ans: In second's clock, it takes 60 s by the seconds hand to rotate by 2π.
So, Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 9: A particle is located at (3 m, 4 m) and moving with Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Find its angular velocity about origin at this instant.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 10: Particle P shown in figure is moving in a circle of radius R = 10cm with linear speed v = 2m/s. Find the angular speed of particle about point O.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

ωc = 2ωo
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 11: Two points P and Q, diametrically opposite on a disc of radius R have linear velocities v and 2v as shown in figure. Find the angular speed of the disc.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ x + 2R=2x
⇒ x=2R
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 12.   Point A of rod AB(l = 2m)is moved upwards against a wall with velocity v = 2m/s. Find angular speed of the rod at an instant when θ = 60°.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Torque
Ques 13: A force Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev is acting on a body at point (2 m, 4 m, -2 m). Find torque of this force about origin.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= i (- 8 + 6) - j (- 4 + 4) + k (6 - 8)
= -2i - 2k = -2(i + k)Nm

Ques 14: A particle of mass m = 1 kg is projected with speed u=20√2m/s at angle θ = 45° with horizontal Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 15: Point C is the centre of mass of the rigid body shown in figure. Find the total torque acting on the body about point C.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: τ = τ10 + τ20 + τ30
= 0 + 20 sin 45° × 0.1 + 30 sin 60° × 0.05
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 16: Find the net torque on the wheel in figure about the point O if a = 10 cm and b = 25 cm.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: τ = 12 cos 60° × 0.1 - 10 × 0.25 - 9 × 0.25
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev


Rotation of a Rigid Body About a Fixed Axis Uniform angular acceleration
Ques 17: A wheel rotating with uniform angular acceleration covers 50 rev in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds.
Ans: Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

= 8π rad/s2
w = αt = 8π rad/s2 × 5s = 40 rad/s

Ques 18: A wheel starting from rest is uniformly accelerated with a = 2 rad/s2 for 5 s. It is then allowed to rotate uniformly for the next two seconds and is finally brought to rest in the next 5 s. Find the total angle rotated by the wheel.
Ans: θ = θ1 + θ2 + θ3
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

α t1 (t1 + t2) = 2 × 5(5 + 2) = 70rad

Ques 19: A wheel whose moment of inertia is 0.03 kg m2, is accelerated from rest to 20 rad/s in 5 s. When the external torque is removed, the wheel stops in 1 min. Find:
(a) the frictional torque, 
(b) the external torque
Ans: ω1 = α1t1
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

is the torque due to friction. Again,
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
or         τ1 = 0.01 + 0.03 × 4 = 0.13 Nm

Ques 20: A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2rad/s2. At what time will the body have kinetic energy same as the initial value if the torque continues to act ?
Ans: ω = ω0 + αt
⇒  - 20 rad/s
= 20 rad/s - 2 rad/s2 × t
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 21: A uniform disc of mass 20 kg and radius 0.5 m can turn about a smooth axis through its centre and perpendicular to the disc. A constant torque is applied to the disc for 3 s from rest and the angular velocity at the end of that time is 240/π rev/min. Find the magnitude of the torque. If the torque is then removed and the disc is brought to rest in t seconds by a constant force of 10 N applied tangentially at a point on the rim of the disc, find t.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Q 22.   A uniform disc of mass m and radius R is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity w0. It is placed on a rough horizontal plane with the axis of the disc keeping vertical. Coefficient of friction between the disc and the surface is m. 
Find:
(a) the time when disc stops rotating,
(b) the angle rotated by the disc before stopping.
Ans:
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) dτ =μ.dm.g.x Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Non-uniform angular acceleration
Ques 23:A flywheel whose moment of inertia about its axis of rotation is 16 kg-m2 is rotating freely in its own plane about a smooth axis through its centre. Its angular velocity is 9 rad s-1 when a torque is applied to bring it to rest in t0 seconds. Find t0 if:
(a) the torque is constant and of magnitude of 4 Nm,
(b) the magnitude of the torque after t seconds is given by kt.
Ans: 

(a) τ = Iα and 0 = ω0 - αt
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ques 24: A shaft is turning at 65 rad/s at time zero. Thereafter, angular acceleration is given by α = -10 rad/s2 - 5t rad/s2where t is the elapsed time.
(a) Find its angular speed at t = 3.0s.
(b) How far does it turn in these 3s ?
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
at t = 0, ω= c = 65 rad/s
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒   dθ  = ωdt
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 25: The angular velocity of a gear is controlled according to ω = 12 - 3t2 where ω, in radian per second, is positive in the clockwise sense and t is the time in seconds. Find the net angular displacement Δθ from the time t = 0 to t = 3 s. Also, find the number of revolutions N through which the gear turns during the 3s.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Δθ (2s) = 12 × 2 - 23 =16rad
θ3 - θ2 = 27 - 16 = 9 rad
So, in third second, N'=Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 26: A solid body rotates about a stationary axis according to the law q = at - bt3, where a = 6 rad/s and b = 2 rad/s3. Find the mean values of the angular velocity and acceleration over the time interval between t = 0 and the time, when the body comes to rest.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ ω = 0 at t = 1s

∴   θ(0) = 0 and θ(1) = 6 - 2 = 4 rad
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= -6 rad / s2


Angular Momentum

Ques 27: A particle of mass 1 kg is moving along a straight line y = x + 4. Both x and y are in metres. Velocity of the particle is 2 m/s. Find magnitude of angular momentum of the particle about origin.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
L = mv cos θ r, tan θ =dy/dx= 1
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 4√2 kg-m2/s

Ques 28: A uniform rod of mass m is rotated about an axis passing through point O as shown. Find angular momentum of the rod about rotational axis.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 29: A solid sphere of mass m and radius R is rolling without slipping as shown in figure. Find angular momentum of the sphere about z-axis.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: L = mvr + Iω
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 30: A rod of mass m and length 2R is fixed along the diameter of a ring of same mass m and radius R as shown in figure. The combined body is rolling without slipping along x-axis. Find the angular momentum about z-axis.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: L = mvr + Iω
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Conservation of Angular Momentum
Ques 31: If radius of earth is increased, without change in its mass, will the length of day increase, decrease or remain same?
Ans: As Iω = constant ⇒ T ∝ I.
With increase in R, I increases and so is T. So, length of day will increase.

Ques 32: The figure shows a thin ring of mass M = 1 kg and radius R = 0.4 m spinning about a vertical diameter. Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
A small bead of mass m = 0.2 kg can slide without friction along the ring. When the bead is at the top of the ring, the angular velocity is 5 rad/s. What is the angular velocity when the bead slips halfway to θ = 45°.

Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: I1ω1 = I2ω2 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 33: A horizontal disc rotating freely about a vertical axis makes 100 rpm. A small piece of wax of mass 10 g falls vertically on the disc and adheres to it at a distance of 9 cm from the axis. If the number of revolutions per minute is thereby reduced to 90. Calculate the moment of inertia of disc.
Ans: 

I1ω1 = (I1 + mr22
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 729 × 10-4 kg-m2

Q 34.   A man stands at the centre of a circular platform holding his arms extended horizontally with 4 kg block in each hand. He is set rotating about a vertical axis at 0.5 rev/s. The moment of inertia of the man plus platform is 1.6 kg-m2, assumed constant. The blocks are 90 cm from the axis of rotation. He now pulls the blocks in toward his body until they are 15 cm from the axis of rotation. 
Find (a) his new angular velocity and (b) the initial and final kinetic energy of the man and platform, (c) how much work must the man do to pull in the blocks ?
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(c) ΔW = E-Ei = 181J - 39.9J = 141.1J

Ques 35: A horizontally oriented uniform disc of mass M and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass m. A light thread running down through the hollow axle of the disc is tied to the body. Initially the body was located at the edge of the disc and the whole system rotated with an angular velocity w0. Then, by means of a force F applied to the lower end of the thread the body was slowly pulled to the rotation axis. Find:
(a) the angular velocity of the system in its final state,
(b) the work performed by the force F.
Ans:
(a) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
W = E- Ei
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev


Pyre Rolling

Ques 36: Consider a cylinder of mass M and radius R lying on a rough horizontal plane. It has a plank lying on its top as shown in figure. A force F is applied on the plank such that the plank moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact. Calculate the acceleration of the cylinder and the frictional forces at the two contacts.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

F cos θ - f1 = m.2a ...(i)
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

While, f+ f= Ma ...(iii)
From Eqs. (ii) and (iii),

Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 37: Find the acceleration of the cylinder of mass m and radius R and that of plank of mass M placed on smooth surface if pulled with a force F as shown in figure. Given that sufficient friction is present between cylinder and the plank surface to prevent sliding of cylinder.
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ra - a = a ⇒ Rα = 2a ⇒ f = ma

∴ F - 2ma = (M + m) a ⇒ F = (M + 3m) a
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 38: In the figure shown a force F is applied at the top of a disc of mass 4 kg and radius 0.25 m. Find maximum value of F for no slipping
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
F + f = ma       ...(i)
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

fmax = μmg = 1/4ma ⇒ a = 4μg
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
3μmg = 3 × 0.6 × 4 × 10 = 72N

Q 39.   In the figure shown a solid sphere of mass 4 kg and radius 0.25 m is placed on a rough surface. Find : (g = 10 m/s2)
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

(a) minimum coefficient of friction for pure rolling to take place.
(b) If μ > μmin, find linear acceleration of sphere.
(c) If Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev, find linear acceleration of cylinder. Here, μmin is the value obtained in part (a).
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
mg sin θ - f = ma ⇒ mg sin θ = 7/5 ma
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) For minimum value of μ,
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Angular Impulse
Ques 40: A uniform rod AB of length 2l and mass m is rotating in a horizontal plane about a vertical axis through A, with angular velocity ω, when the mid-point of the rod strikes a fixed nail and is brought immediately to rest. Find the impulse exerted by the nail.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 41: A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod.
(a) What is the final angular velocity of the rod ?
(b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision ?
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 42: A uniform rod AB of mass 3m and length 2l is lying at rest on a smooth horizontal table with a smooth vertical axis through the end A. A particle of mass 2m moves with speed 2m across the table and strikes the rod at its mid-point C. If the impact is perfectly elastic. Find the speed of the particle after impact if:
(a) it strikes the rod normally,
(b) its path before impact was inclined at 60° to AC.
Ans: 
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(a) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
where, 2u= lω + v
⇒ v = -lω + 2u 4ml2
ω =2m(4u - lω)l
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics

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Solution by DC Pandey NEET Notes | EduRev

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Mechanics of Rotational Motion: JEE Main (Part- 1) - Physics

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Solution by DC Pandey NEET Notes | EduRev

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