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Paragraph Based Questions
(A) Three bodies A(ring) ,B(Solid cylinder) ,C (solid sphere) rolls down the same inclined plane of angle $\alpha$ and height H with out slipping.
The radii of the bodies are identical and equal to R.The mass of the bodies are
M_{1}, M_{2}, M_{3} respectively. The relation in the masses is given as below
M_{1} > M_{2} > M_{3}
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) 2v_{cm}
(c) v_{cm}
(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) 2v_{cm}
(c) v_{cm}
(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 Solution 1-6
Kinetic energy of the Rolling body is given by
$K=\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)$
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,
$mgH=\frac{1}{2}mv^2\left(1+\frac{k^2}{R^2}\right)$
Or
$v^2=\left(\frac{2gH}{1+\frac{k^2}{R^2}}\right)$
So velocity of center of mass is independent of the mass of the body
Now for Ring
$k^2=R^2$
So
$v_{ring}=\sqrt{gH}$
Now for Solid Cylinder
$k^2=R^2/2$
So
$v_{solid}=\sqrt{\frac{4gH}{3}}$
Now of Solid Sphere
$k^2=2R^2/5$
So
$v_{sphere}=\sqrt{\frac{10gH}{7}}$
So it is clear that
$v_{sphere} > v_{solid} > v_{ring}$
now also
$\omega=\frac {v}{R}$
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) $I=\frac{mL^2}{144}$
(b) $I=\frac{14mL^2}{144}$
(c) $I=\frac{151mL^2}{144}$
(d) $I=\frac{149mL^2}{144}$ Solution
Question 8
Find the angular acceleration of the rod at the instant it is release
(a) $\alpha=\frac{348g}{14L}$
(b) $\alpha=\frac{348g}{19L}$
(c) $\alpha=\frac{34g}{149L}$
(d) $\alpha=\frac{348g}{149L}$ Solution
$\tau=g(\frac{mL}{4}+\frac{2mL}{3}+\frac{3mL}{2})$
$\tau=\frac{29mLg}{12}$
Now
$\tau=I\alpha$
Substituting the values
$\alpha=\frac{348g}{149L}$
(C) A Projectile of mass m is fired from point O(origin) with an initial velocity u at angle Î¸ 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,$\frac{-m(ucos{\theta})gt^2}{2}$
(b) 0,0,$\frac{m(ucos{\theta})gt^2}{2}$
(c) 0,$u cos \theta$,$\frac{-m(ucos{\theta})gt^2}{2}$
(d) 0,$u cos \theta$,$\frac{m(ucos{\theta})gt^2}{2}$ Solution
Answer is (a)
Displacement components at time t
$x=(ucos{\theta})t$
$y=(usin{\theta})t-\frac{1}{2}gt^2$
$z=0$
Velocity components at time t
$v_x=(ucos{\theta})$
$v_y=(usin{\theta})-gt$
$v_z=0$
Now angular momentum about origin
$\mathbf{L}=\mathbf{r} \times m\mathbf{v}$
$\mathbf{L}=m(x\mathbf{i}+y\mathbf{j}) \times (v_x\mathbf{i}+v_y\mathbf{j})$
$\mathbf{L}=m(xv_y-yv_x)\mathbf{k}$
Substituting the values from displacement and velocity in above equation
$\mathbf{L}=\frac{-m(ucos{\theta})gt^2\mathbf{k}}{2}$
Question 10
Calculate the torque of the weight acting on the projectile about the Origin
(a)$mgu (cos \theta) t\mathbf{k}$
(b) $mgu (cos \theta) t\mathbf{j}$
(c) $-mgu (cos \theta) t\mathbf{k}$
(d) None of the above Solution
(D) A horizontal turn table carries a gun at point X and rotates with initial angular velocity Ï‰_{0}Â about its vertical geometric axis. The Gun fires a bullet of mass m with tangential muzzle velocity v
I_{0} 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) $I_0\omega+mr^2\omega$
(b) $I_0\omega$
(c) $I_0\omega-mr^2\omega$
(d) None of these Solution
Answer is (a)
Initial angular momentum of the system
= angular momentum of the (turntable + gun) + angular momentum of the bullet
$=I_0\omega+mr^2\omega$
Question 12
Find the final angular velocity of the turntable
(a)$\omega_0+\frac{mvr}{I_0}$
(b) $\omega_0$
(c) $\omega_0-\frac{mvr}{I_0+mr^2}$
(d) $\omega_0+\frac{mvr}{I_0+mr^2}$ Solution
Answer is (d)
Let $\omega$ be the angular velocity after the firing
The absolute velocity of the bullet to the right
$=v-r\omega$
Final angular momentum of the system
$=I_0\omega-m(v-r\omega)r$
Since external torque is zero about axis of rotation, angular momentum will be conserved about the axis of rotation
$I_0\omega+mr^2\omega=I_0\omega-m(v-r\omega)r$
Or
$\omega=\omega_0+\frac{mvr}{I_0+mr^2}$
(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) $\frac{mv^2}{3}$
(b) $\frac{mv^2}{2}$
(c) $\frac{mv^2}{4}$
(d) $\frac{mv^2}{6}$ 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
$\frac{2mL^2}{9}$
(c) Net torque and net force is zero on the system
(d) None of these Solution 13-18
Let us assume
$u$-> velocity of the center of mass of the system after the collision
$\omega$ -> 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
$x_0=\frac{2M \times L/2+M \times 0}{2M+M}=\frac{L}{3}x$
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
$=\frac{2ML^2}{12}=\frac{ML^2}{6}$
Now using parallel axis theorem, moment of inertia about the center of mass the system
$=\frac{ML^2}{6}+2M \times \left(\frac{L}{6}\right)^2=\frac{2ML^2}{9}$
The point mass will add $M(\frac{L}{3})^2$ to the total
So
$I_{system}=\frac{2ML^2}{9}+\frac{ML^2}{9}=\frac{ML^2}{3}$
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
$\frac{MvL}{3}=I\omega=\frac{ML^2}{3}\omega$
Or
$\omega=v/L$
Now we know the values of u and $\omega$
Velocity of Point A
$=u+\omega \times \frac{L}{3}$
$=\frac {2v}{3}$
Velocity of point B
$=u-\omega \times \frac{2L}{3}$
$=\frac {-v}{3}$
Velocity of point C
$=u-\omega \times \frac{L}{3}$
=0
Velocity of Point D
$=u-\omega(\frac{L}{2}-\frac{L}{3})$
=v/6
Let check the KE before and after collision
$KE_i=\frac{1}{2}Mv^2$
$KE_f=\frac{1}{2}(3M)u^2+\frac{1}{2}I\omega^2$
Substituting the values from above
$KE_f=\frac{1}{3}Mv^2$
Since $KE_i > KE_f$ , 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) $\frac{mvh}{mh^2+ML^2}$
(b) $\frac{mvh}{mh^2+\frac{1}{3}ML^2}$
(c) $\frac{mvh}{mh^2+\frac{1}{2}ML^2}$
(d) None of these Solution
Angular momentum is conserved around A as no external torque is acting on the system(Rod + bullet)
Initial Angular Momentum =mvh
Final Angular Momentum= IÏ‰
$=(mh^2+\frac{1}{3}ML^2)\omega$
So $\omega=\frac{mvh}{mh^2+\frac{1}{3}ML^2}$
Question 20 Find the Linear Momentum of the system just after the collision if Ï‰ is the angular velocity just after the collision
(a)(mh+ML/2) Ï‰
(b) (mh+ML) Ï‰
(c) (mh+ML/3) Ï‰
(d) None of these Solution
Total Linear momentum =Linear momentum of mass m + Linear momentum of the center of mass of the rod
Now $v=r \omega$
So $v_m=h \omega$
and $v_{cm}=(L/2) \omega$
So Total linear momentum
=mh Ï‰ + M(L/2) Ï‰
=(mh+ML/2)Ï‰
(G)An upright hoop is projected on a pavement with a initial horizontal speed v_{0} 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 Î¼ be the coefficient of friction Question 21 Find the time taken by the hoop to start rolling with out slipping
(a) v_{0}/2 Î¼g
(b) v_{0}/2 g
(c) v_{0}/ Î¼g
(d) None of these Solution
Let f be the frictional force acting
Then
$-f=Ma$
So a=-f/M
Now
$v=u+at$
So
$v=v_0 – \frac {ft}{M}$
Now
Torque of frictional force
=fR
Now we know that
$\tau=T\alpha$
Substituting the values
$fR=I \alpha$
or $ \alpha =\frac {f}{MR}$ as $I=MR^2$
Now we know that
$\omega=\omega_0+\alpha t$
As intial angular velocity is zero
$ \omega =\frac {ft}{MR}$
Now for no slipping condition
$v=R \omega$
$(v_0-\frac{ft}{M})=R(\frac{ft}{MR})$
Or $v_0= \frac {2ft}{M}$
Or $t=\frac {v_0M}{2f}$
Now Frictional force is given by
$f= \mu Mg$
So
$t=\frac {v_0}{2 \mu g}$
So answer is (a)
Question 22 Find the velocity of the hoop at no slipping
(a) v_{0}/2
(b) 3 v_{0}/2
(c) v_{0}/4
(d) None of these Solution
$v=v_0 – \frac {ft}{M}$
Substituting the values of f and t
We get
$v= \frac {v_0}{2}$
Question 23 which one of the following statement are correct
(a) The loop will have total energy $\frac{Mv_0^2}{4}$ when it start pure rolling
(b) $\frac{3Mv_0^2}{4}$ Energy is lost in friction during the whole motion
(c) The velocity of the top most point is v_{0} when it start pure rolling
(d) $\frac{Mv_0^2}{8}$ is the rotational energy of the loop when it start pure rolling Solution
All are correct
Total energy at the starting =$\frac{1}{2}Mv_0^2$
Total energy at pure rolling stage
=Translational KE + Rotational KE
$=\frac{1}{2}M\left(\frac{v_0}{2}\right)^2+\frac{1}{2}I\omega^2$
$=\frac{Mv_0^2}{8}+\frac{1}{2}MR^2\left(\frac{v_0}{2R}\right)^2$
$=\frac{Mv_0^2}{4}$
So loss in Energy
$=\frac{3Mv_0^2}{4}$
(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)$\frac{2}{3}a$
(b)$\frac{1}{3}a$
(c)$\frac{1}{2}a$
(d)$\frac{1}{4}a$ Solution
Answer is (b)
Let $a_1$ 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
$f-F_{fict}=Ma_1$ ---(1)
Let $\alpha$ be the angular acceleration ,Then torque equation
$Rf=-I\alpha$ --(2)
Negative sign is present as torque and angular acceleration are opposite
Also
$F_{fict}=Ma$
$I=\frac{MR^2}{2}$ ----(3)
Solving equation (1),(2) and (3)
We get
$a_1=\frac{-2}{3}a$
So acceleration with respect to Ground
$=-\frac{2}{3}a+a=\frac{1}{3}a$
Question 25 Find the frictional force acting on the cylinder
(a)$\frac{Ma}{3}$
(b)$\frac{Ma}{2}$
(c)$\frac{2Ma}{3}$
(d) $\frac{Ma}{4}$ Solution
Answer is (a)
From Ground frame of reference, frictional force is the only force acting on the cylinder
So
$f=\frac{Ma}{3}$
Question 26 Find the angular acceleration of the cylinder about the axis passing through center of mass
(a) $\frac{a}{3R}$
(b) $\frac{a}{4R}$
(c) $\frac{a}{2R}$
(d) $\frac{2a}{3R}$ Solution
Answer is (d)
For the point of cylinder which comes in contact with plank, the acceleration should be should be same as plank acceleration
$\frac{1}{3}a+R\alpha=a$
Or
$\alpha=\frac{2a}{3R}$
Multiple Choice Questions
Question 27
Position vector of a body of mass 6 Kg is given as
$\mathbf{r}=\mathbf{i}(3t^2-6t)+\mathbf{j}(-4t^3)$
Find the torque acting on it about the origin
(a) $\mathbf{k}(-288t^3+864t^2)$
(b)$\mathbf{i}(-288t^3+864t^2)$
(c) $\mathbf{j}(-288t^3+864t^2)$
(d) None of these Solution
$\mathbf{r}=\mathbf{i}(3t^2-6t)+\mathbf{j}(-4t^3)$
Velocity vector
$\frac{d\mathbf{r}}{dt}=\mathbf{i}(6t-6)+\mathbf{j}(-12t^2)$
Acceleration vector
$\frac{d^2\mathbf{r}}{dt^2}=6\mathbf{i}-24t\mathbf{j}$
Net Force vector
$\mathbf{F}=m\mathbf{a}=6(6\mathbf{i}-24t\mathbf{j})$
Now Torque is given
$\overrightarrow{\tau}=\mathbf{r} \times\mathbf{F}$
$=[\mathbf{i}(3t^2-6t)+\mathbf{j}(-4t^3)] \times [6(6\mathbf{i}-24t\mathbf{j})]$
$=\mathbf{k}(-288t^3+864t^2)$
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 $\frac{5g}{11a}$
(b) Moment of inertia of the system about pivot is 11ma^{2}
(c) The angular acceleration at the instant when the rod makes an angle Î¸ with horizontal $\frac{5gcos{\theta}}{11a}$
(d) The net torque about the pivot end becomes zero when rod becomes vertical. Solution
All are correct
Total torque about the pivot in horizontal position
$\tau=g(2ma+3ma)=5mga$
Moment of Inertial about the pivot end
$I=2ma^2+m(3a)^2=11ma^2$
Now
$\tau=I\alpha$
Or
$\alpha=\frac{\tau}{I}$
So
$\alpha=\frac{5g}{11a}$
Total torque about the pivot when the rod makes angle $\theta$ with horizontal
$\tau=g(2macos{\theta}+3macos{\theta})=5mgacos{\theta}$
So angular acceleration becomes
$\alpha=\frac{5gcos{\theta}}{11a}$
When $\theta=90$, torque becomes zero
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