In this page we have Important Objective questions on rotation for JEE Main/Advanced,CBSE,AP Physics . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Paragraph Based Questions
(A) Three bodies A(ring) ,B(Solid cylinder) ,C (solid sphere) rolls down the same inclined plane of angle α and height H with out slipping.
The radii of the bodies are identical and equal to R.The mass of the bodies are
M1, M2, M3 respectively. The relation in the masses is given as below
M1 > M2 > M3
Question 1
Statement I: The velocity of the center of mass of the bodies at the bottom of the incline plane is dependent on the mass of the bodies
Statement II: The total kinetic energy( Translation + Rotational) will be largest for the body A at the bottom of the inclined plane
(a) Statement I is true only
(b) Statement II is true only
(c) Both the statements are true
(d) Both of them are false Question 2
Which one will have largest Center of mass velocity at the bottom of the incline plane
(a) A
(b) B
(c) C
(d) Data insufficient Question 3
Which one will have least center of mass velocity at the bottom of the incline plane
(a) A
(b) B
(c) C
(d) Data insufficient Question 4
At any time during the motion, the velocity of the point in contact of the body with the incline plane
(a) 0
(b) 2vcm
(c) vcm
(d) None of the above Question 5
At any time during the motion, the velocity of the uppermost point of the body
(a) 0
(b) 2vcm
(c) vcm
(d) None of the above Question 6
The angular velocity will be maximum at the bottom of the incline place for the body
(a) A
(b) B
(c) C
(d) Data insufficient
Answer(1-6)
Kinetic energy of the Rolling body is given by K=12mv2(1+k2R2)
Where k is the radius of gyration of the body and v is the velocity of the center of mass of the body
Now from the law of conservation of energy
Potential energy lost=Kinetic energy gained
So, mgH=12mv2(1+k2R2)
Or v2=(2gH1+k2R2)
So velocity of center of mass is independent of the mass of the body
Now for Ring k2=R2
So vring=√gH
Now for Solid Cylinder k2=R2/2
So vsolid=√4gH3
Now of Solid Sphere k2=2R2/5
So vsphere=√10gH7
So it is clear that vsphere>vsolid>vring
now also ω=vR
So Angular velocity will also be same
Hence the Answers are
(1) d
(2) c
(3) a
(4) a
(5) b
(6) c
(B)A nearly mass less rod (AB) of Length L is pivoted at one end A so that it can swing free as a pendulum. Three masses A(m) ,B(2m) and C( 3m) are attached to it at the distance L/4,L/3 and L/2 respectively from end A. The rod is held horizontal and then it is released
Question 7
Find the Moment of the Inertia of system about the pivoted end
(a) I=mL2144
(b) I=14mL2144
(c) I=151mL2144
(d) I=149mL2144
Answer
I=m(L4)2+2m(L3)2+3m(L2)2 I=149mL2144
Question 8
Find the angular acceleration of the rod at the instant it is release
(a) α=348g14L
(b) α=348g19L
(c) α=34g149L
(d) α=348g149L
Answer
τ=g(mL4+2mL3+3mL2) τ=29mLg12
Now τ=Iα
Substituting the values α=348g149L
(C) A Projectile of mass m is fired from point O(origin) with an initial velocity u at angle θ above the horizontal
Question 9
Find the x, y and z component of the angular momentum of the projectile about point O as a function of time
(a) 0,0,−m(ucosθ)gt22
(b) 0,0,m(ucosθ)gt22
(c) 0,ucosθ,−m(ucosθ)gt22
(d) 0,ucosθ,m(ucosθ)gt22
Answer
Answer is (a)
Displacement components at time t x=(ucosθ)t y=(usinθ)t−12gt2 z=0
Velocity components at time t vx=(ucosθ) vy=(usinθ)−gt vz=0
Now angular momentum about origin L=r×mv L=m(xi+yj)×(vxi+vyj) L=m(xvy−yvx)k
Substituting the values from displacement and velocity in above equation L=−m(ucosθ)gt2k2
Question 10
Calculate the torque of the weight acting on the projectile about the Origin
(a)mgu(cosθ)tk
(b) mgu(cosθ)tj
(c) −mgu(cosθ)tk
(d) None of the above
Answer
→τ=r×F →τ=(xi+yj)×(−mgj) =−mgxk =−mg(ucosθ)tk
(D) A horizontal turn table carries a gun at point X and rotates with initial angular velocity ω0 about its vertical geometric axis. The Gun fires a bullet of mass m with tangential muzzle velocity v
I0 is the moment of inertia of the turntable and gun excluding bullet
Question 11
Find the initial angular momentum of the system about the axis of rotation
(a) I0ω+mr2ω
(b) I0ω
(c) I0ω−mr2ω
(d) None of these
Answer
Answer is (a)
Initial angular momentum of the system
= angular momentum of the (turntable + gun) + angular momentum of the bullet =I0ω+mr2ω
Question 12
Find the final angular velocity of the turntable
(a)ω0+mvrI0
(b) ω0
(c) ω0−mvrI0+mr2
(d) ω0+mvrI0+mr2
Answer
Answer is (d)
Let ω be the angular velocity after the firing
The absolute velocity of the bullet to the right =v−rω
Final angular momentum of the system =I0ω−m(v−rω)r
Since external torque is zero about axis of rotation, angular momentum will be conserved about the axis of rotation I0ω+mr2ω=I0ω−m(v−rω)r
Or ω=ω0+mvrI0+mr2
(E)A uniform rod (AB) of length L and mass 2M rest on smooth horizontal surface. A point mass M moving horizontally at right angles to the rod with initial velocity v collides with one end (A) and sticks to it
Question 13
Find the velocity of the center of mass of the system after the collision
(a) v/3
(b) v/2
(c) v/4
(d) 3v/4 Question 14
Find the angular velocity of the rod after the collision
(a)v/L
(b) v/2L
(c) 2v/L
(d) 3v/4L Question 15 Match the column Column A -> Velocity of the various part of the rod Column B -> Their values in terms of v Column A
(P) Velocity of the end A
(Q) Velocity of the end B
(R) Velocity of the point C which is at distance 2L/3 from point A
(S) Velocity of point D which is at distance L/2 from A Column B
(U) -v/3
(V) 2v/3
(W) 0
(X) v/6 Question 16
which of the following will remain true in collision
(a) Law of conservation of linear momentum
(b) Law of conservation of angular momentum
(c) Law of conservation of energy
(d) None of these Question 17 Find the total energy of the system after the collision
(a) mv23
(b) mv22
(c) mv24
(d) mv26 Question 18 Which of the following is true?
(a) The Center of mass velocity before and after the collision will remain same
(b) Moment of inertia about center of mass of the rod is 2mL29
(c) Net torque and net force is zero on the system
(d) None of these
Answer(13-18)
Let us assume u-> velocity of the center of mass of the system after the collision ω -> angular velocity of the system about center of mass after collision
First we need to find the center of mass of the system
Let's measure the distance from end A
Then x0=2M×L/2+M×02M+M=L3x
Now we also need to find the moment of inertia about center of mass
We know that for rod, moment of inertia about its center of mass =2ML212=ML26
Now using parallel axis theorem, moment of inertia about the center of mass the system =ML26+2M×(L6)2=2ML29
The point mass will add M(L3)2 to the total
So Isystem=2ML29+ML29=ML23
Now since no external force acts, Linear momentum remains conserved
Also since no external torque acting, angular momentum remain conserved also
Applying law of conservation of linear momentum Mv=3Mu
Or
u=v/3
Applying law of conservation of angular momentum about system center of mass MvL3=Iω=ML23ω
Or ω=v/L
Now we know the values of u and ω
Velocity of Point A =u+ω×L3 =2v3
Velocity of point B =u−ω×2L3 =−v3
Velocity of point C =u−ω×L3
=0
Velocity of Point D =u−ω(L2−L3)
=v/6
Let check the KE before and after collision KEi=12Mv2 KEf=12(3M)u2+12Iω2
Substituting the values from above KEf=13Mv2
Since KEi>KEf , Energy is not conserved
Hence the Answers are
(13) (a)
(14) (a)
(15)
P ->V
Q-> U
R-> W
S-> X
(16) (a) and (b)
(17) (d)
(18) (a),(b),(c)
(F)A rod of length L and mass M can freely rotate around a pivot at A. A bullet of mass m and velocity v hit the rod at a height h from A and becomes imbedded in it
Question 19 Find the angular velocity of the Rod just after the collision
(a) mvhmh2+ML2
(b) mvhmh2+13ML2
(c) mvhmh2+12ML2
(d) None of these
Answer
Angular momentum is conserved around A as no external torque is acting on the system(Rod + bullet)
Initial Angular Momentum =mvh
Final Angular Momentum= Iω =(mh2+13ML2)ω
So ω=mvhmh2+13ML2
Question 20 Find the Linear Momentum of the system just after the collision if ω is the angular velocity just after the collision
(a)(mh+ML/2) ω
(b) (mh+ML) ω
(c) (mh+ML/3) ω
(d) None of these
Answer
Total Linear momentum =Linear momentum of mass m + Linear momentum of the center of mass of the rod
Now v=rω
So vm=hω
and vcm=(L/2)ω
So Total linear momentum
=mh ω + M(L/2) ω
=(mh+ML/2)ω
(G)An upright hoop is projected on a pavement with a initial horizontal speed v0 but with out spin so that its slide. The resulting frictional force causes the hoop to spin to loose translational speed and to acquire an angular speed. Eventually the hoop rolls without slipping and let μ be the coefficient of friction Question 21 Find the time taken by the hoop to start rolling with out slipping
(a) v0/2 μg
(b) v0/2 g
(c) v0/ μg
(d) None of these
Answer
Let f be the frictional force acting
Then −f=Ma
So a=-f/M
Now v=u+at
So v=v0ftM
Now
Torque of frictional force
=fR
Now we know that τ=Tα
Substituting the values fR=Iα
or α=fMR as I=MR2
Now we know that ω=ω0+αt
As intial angular velocity is zero ω=ftMR
Now for no slipping condition v=Rω (v0−ftM)=R(ftMR)
Or v0=2ftM
Or t=v0M2f
Now Frictional force is given by f=μMg
So t=v02μg
So answer is (a)
Question 22 Find the velocity of the hoop at no slipping
(a) v0/2
(b) 3 v0/2
(c) v0/4
(d) None of these
Answer
v=v0ftM
Substituting the values of f and t
We get v=v02
Question 23 which one of the following statement are correct
(a) The loop will have total energy Mv204 when it start pure rolling
(b) 3Mv204 Energy is lost in friction during the whole motion
(c) The velocity of the top most point is v0 when it start pure rolling
(d) Mv208 is the rotational energy of the loop when it start pure rolling
Answer
All are correct
Total energy at the starting =12Mv20
Total energy at pure rolling stage
=Translational KE + Rotational KE =12M(v02)2+12Iω2 =Mv208+12MR2(v02R)2 =Mv204
So loss in Energy =3Mv204
(H)A cylinder of Mass M and radius R is kept on the plank. The plank starts moving with acceleration a. The cylinder rolls without slipping on the plank Question 24 Find the acceleration of the cylinder center of mass from the laboratory frame of reference
(a)23a
(b)13a
(c)12a
(d)14a
Answer
Answer is (b)
Let a1 be the acceleration of the cylinder center of mass with respect to frame fixed to plank.
Let f be the frictional force acting on the cylinder
Now from Force equation f−Ffict=Ma1 ---(1)
Let α be the angular acceleration ,Then torque equation Rf=−Iα --(2)
Negative sign is present as torque and angular acceleration are opposite
Also Ffict=Ma I=MR22 ----(3)
Solving equation (1),(2) and (3)
We get a1=−23a
So acceleration with respect to Ground =−23a+a=13a
Question 25 Find the frictional force acting on the cylinder
(a)Ma3
(b)Ma2
(c)2Ma3
(d) Ma4
Answer
Answer is (a)
From Ground frame of reference, frictional force is the only force acting on the cylinder
So f=Ma3
Question 26 Find the angular acceleration of the cylinder about the axis passing through center of mass
(a) a3R
(b) a4R
(c) a2R
(d) 2a3R
Answer
Answer is (d)
For the point of cylinder which comes in contact with plank, the acceleration should be should be same as plank acceleration 13a+Rα=a
Or α=2a3R
Multiple Choice Questions
Question 27
Position vector of a body of mass 6 Kg is given as r=i(3t2−6t)+j(−4t3)
Find the torque acting on it about the origin
(a) k(−288t3+864t2)
(b)i(−288t3+864t2)
(c) j(−288t3+864t2)
(d) None of these
Answer
r=i(3t2−6t)+j(−4t3)
Velocity vector drdt=i(6t−6)+j(−12t2)
Acceleration vector d2rdt2=6i−24tj
Net Force vector F=ma=6(6i−24tj)
Now Torque is given →τ=r×F =[i(3t2−6t)+j(−4t3)]×[6(6i−24tj)] =k(−288t3+864t2)
Question 28
A nearly massless rod is pivoted at one end so it can freely swing as a pendulum. Two masses 2m and m are attached to it at distance a and 3a respectively from the pivot end. The rod is held horizontal and then released.
Which of the following is correct?
(a) The angular acceleration at the instant it is released 5g11a
(b) Moment of inertia of the system about pivot is 11ma2
(c) The angular acceleration at the instant when the rod makes an angle θ with horizontal 5gcosθ11a
(d) The net torque about the pivot end becomes zero when rod becomes vertical.
Answer
All are correct
Total torque about the pivot in horizontal position τ=g(2ma+3ma)=5mga
Moment of Inertial about the pivot end I=2ma2+m(3a)2=11ma2
Now τ=Iα
Or α=τI
So α=5g11a
Total torque about the pivot when the rod makes angle θ with horizontal τ=g(2macosθ+3macosθ)=5mgacosθ
So angular acceleration becomes α=5gcosθ11a
When θ=90, torque becomes zero