Question 1

The  x  location and mass of two masses are given as  \left(1m, 3kg\right)  and  \left(-2m, 2kg\right) . A third mass weighs  1kg .Given that the center of mass of the system lies at  x=-1m , what must be the  x  position of the third mass?

Accepted Answer

Given :     x_1 = 1  m        m_1  = 3  kg               m_2  =2kg        x_2  = -2  m               m_3 = 1  kgPosition of centre of mass         X = -1  mUsing      X = \dfrac{m_1x_1 + m_2x_2 +m_3x_3}{m_1+m_2+m_3} 	herefore          -1 = \dfrac{3	imes 1 + 2	imes (-2) + 1	imes x_3}{3+2+1}                        \implies x_3 = -5  m

Question 2

Find the position of center of mass of the uniform lamina shown in figure.

Accepted Answer

Answer is C. A_1  = area of complete circle =  \pi\, a^2 A_2  = area of small circle =  \pi \left (\displaystyle \frac {a}{2} \right )^2\, =\, \displaystyle \frac {\pi\, a^2}{4} (x_1,\,y_1)  = coordinates of center of mass of large circle = (0, 0) (x_2,\,y_2)  = coordinates of center of mass of small circle =  \left (\displaystyle \frac {a}{2},\, 0 \right ) Using  x_{COM}\, =\, \displaystyle \frac {A_1x_1\, -\, A_2x_2}{A_1\, -\, A_2} we get  x_{COM}\, =\, \displaystyle \frac {-\displaystyle \frac {\pi a^2}{4} \left ( \displaystyle \frac {a}{2} \right )}{\pi a^2\, -\, \displaystyle \frac {\pi^2}{4}}\, =\, \displaystyle \frac {-\left ( \displaystyle \frac {1}{8} \right )}{\left (\displaystyle \frac {3}{4} \right )}\, a\, =\, -\, \displaystyle \frac {a}{6} and get  x_{COM}\, =\, 0  as  y_1  and  y_1  both are zero.Therefore, coordinates of COM of the lamina shown in figure are  \left ( - \displaystyle \frac {a}{6},\,0 \right ) .

Question 3

A rod of length  L  has a linear mass density   \lambda = Ax + B  ( A  and  B  are constants). Determine the location of the centre of mass from one end of the rod.

Accepted Answer

\displaystyle X_{cm} = \dfrac{\int^{L}_{0} x\ dm}{\int^{L}_{0} dm}    dm = \lambda\ dx = (Ax + B) dx    X_{cm} = \dfrac{\int^{L}_{0} x. (Ax + B)\ dx}{\int^{L}_{0} (Ax + B) dx}     = \dfrac{\dfrac{AL^3}{3} + \dfrac{BL^2}{2}}{\dfrac{AL^2}{2} + BL}    \Rightarrow X_{cm} = \dfrac{L (2AL + 3B)}{3(AL + 2B)}   Correct Option - (C)

Question 4

A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg.If he starts running and when he reaches the other end , he has a velocity  2ms^{-1}  with respect to the boat.The final velocity of the boat is (in  ms^{-1} ).

Accepted Answer

Let final velocity of boat be  v .So velocity of man with respect to ground is  2+v As Center-of-mass of system should remain at rest, in absence of external force,  50(2+v) + 200v = 0\implies v=-\dfrac{2}{5}m/s negative sign idicates that the velocity of boat is in opposite direction to that of the man.

Question 5

Two men 'A' and 'B' are standing on a plank. 'B' is at the middle of the plank and 'A' is at the left end of the plank. The lower surface of the plank is smooth. The system is initially at rest and masses are as shown in the figure. 'A' and 'B' start moving such that the position of 'B' remains fixed with respect to ground. A and B meets at some position. Then the point where A meets B is located at

Accepted Answer

The position of B will remian fixed wrt to ground So A has to move 60 cm along B Lets assume planks moves x distance along leftwardSo  m_A	imes 60=m_P	imes x 40	imes 60=40x x=60m  , so A and B will meet at the right end of plank

Question 6

A body A of mass M falling vertically downwards under gravity breaks into two part; a body B of mass  \dfrac{M}{3}  and a body C of mass  \dfrac{2}{3}  M. The COM of bodies B and C taken together shifts compared to that of body A towards :

Accepted Answer

By Newtons second law for system of bodies,  { F }_{ external }=m{ a }_{ CM } . This equation does not involve internal forces between bodies. SInce in this case, no horizontal force exists on the system of two parts and hence COM will not shift.

Question 7

A disc of radius  R   and mass  M  is cut out from a large disc of radius  2R  in such a way that the edge of the hole touches the edge of the disc. Then moment of inertia of  residual disc about centre is:

Accepted Answer

We can assume cavity part as negative mass.so whole system becomes disc of radius R with positive mass  +  disc with radius R with negative mass. mass of small disc is  M_R=-M  so mass of bigger disc would be  M_{2R}=4M Moment of inertia for bigger disc about axis  I_1=\dfrac{4M(2R)^2}{2}={8MR^2} Moment of inertia for smaller disc about its centre of mass I_2=\dfrac{-MR^2}{2} Moment of inertia for bigger disc about its axis using parallel axis theorem I_2=\dfrac{-MR^2}{2}-MR^2=-\dfrac{3MR^2}{2} Total Moment of inertia about axis  I=I_1+I_2 I= 8MR^2 -\dfrac{3MR^2}{2}=\dfrac{13MR^2}{2} I= \dfrac{13MR^2}{2}

Question 8

Three point sized bodies each of mass  M  are fixed at three corners of a light triangular frame of side length  L . About an axis perpendicular to the plane of frame and passing through centre of frame the moment of inertia of three bodies is:

Accepted Answer

Height of the triangle is,  h = \dfrac{ \sqrt{3}}{2}L The center  (centroid)  of the frame is at  \dfrac{2}{3}h= \dfrac{2}{3} \dfrac{ \sqrt{3}}{2}L=\dfrac{1}{\sqrt{3}}L  from each of the vertex.MI of the system about  center of the frame is:  I = 3 	imes ( M (\dfrac{1}{\sqrt{3}}L)^2)= ML^2

Question 9

Surface mass density of a disc of mass  m  and radius  R  is  \sigma = Kr^2 . then its moment of inertia w.r.t. axis of rotation passing through centre and perpendicular to the plane of disc is:

Accepted Answer

dI = dmr^2 I = \displaystyle \int_0^R Kr^2 2\pi \, rdr . r^2 = K2 \pi \dfrac{R^6}{6} = \dfrac{K \pi R^6}{3} \displaystyle \int_0^m dm = \displaystyle \int_0^R Kr^2 2 \pi rdr m = 2 \pi K \dfrac{R^4 }{4} = \dfrac{\pi KR^4}{2}  I = \dfrac{2}{3} mR^2

Question 10

Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.

Accepted Answer

Since we have a compound object in both cases, we can use the parallel axis theorem to find the moment of inertia about each axis. In (a). the center of mass of the sphere is located at a distance  L + R  from the axis of rotation. In (b), the center of mass of the sphere is located at a distance R from the axis of rotation. In both cases, the moment of inertia of the rod Is about an axis at one end. Refer to (Figure) for the moments of Inertia for the Individual objects. 1.  I_{total} = \sum I_i = I_{Rod} + I_{Sphere};  I_{Sphere} = I_{center \, of \, mass} + m_{Sphere} (L + R)^2 = \dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} (L + R)^2 ;  I_{total} = I_{Rod} + I_{Sphere} = \dfrac{1}{3} m_{Rod} L^2 + \dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} (L + R)^2 ;  I_{total} = \dfrac{1}{3} (2.0 kg) (0.5 m)^2 + \dfrac{2}{5} (1.0 kg)(2.0 m)^2 + (1.0 kg)(0.5 m + 0.2 m)^2;  I_{total} = (0.167 + 0.016 + 0.490) kg . m^2 = 0.673 kg. m^2 .2.  I_{Sphere} = \dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} R^2  I_{total} = I_{Rod} + I_{Sphere} = \dfrac{1}{3} m_{Rod} L^2 + \dfrac{2}{5} m_{Sphere} R^2 + m_{Sphere} R^2; I_{total} = \dfrac{1}{3} (2.0 kg) (0.5 m)^2 + \dfrac{2}{5} (1.0 kg)(0.2 m)^2 + (1.0 kg) (0.2 m)^2 ; I_{total} = (0.167 + 0.016 + 0.04) kg . m^2 = 0.223 kg . m^2 .

Question 11

A solid sphere of radius  R  is set into motion on a rough horizontal surface with a linear speed  {v}_{0}  in forward direction and an angular velocity  { \omega  }_{ 0 }={ v }_{ 0 }/2R  in counter coefficient of friction is  \mu , then find the time after which sphere starts pure rolling.

Accepted Answer

The point of contact of sphere has non-zero velocity w.r.t. ground in the forward direction, frictional force will apply in the backward direction. This would decrease both the linear velocity and the angular velocity. frictional force is  f=\mu mg .Linear deceleration is  a = f/m = \mu g Let the time at which pure rolling starts be  t .So the linear velocity at this time gives by,  v_t = v_o - at = v_o - \mu gt Angular acceleration is  \alpha = \dfrac{fR}{2mR^2/5} = 5\mu g/2R At time  t , the angular velocity is given by, \omega_t = -\omega_o + \alpha t = -v_o/2R + 5\mu gt/2R We have taken angular velocity as negative here to indicate the reverse spin, and angular acceleration positive as it is in the forward spin direction.For pure rolling,  v_{t} = \omega_{t}R\implies v_o - \mu gt = -v_o/2 + 5\mu gt/2\implies3v_0/2 = 7\mu gt/2\implies t = 3v_o/7\mu g

Question 12

When a cylinder is spinned and placed somewhere on surface of a rough inclined plane as shown then after release.

Accepted Answer

When the cylinder is spinning and coming down, the two main forces acting on the ball are gravity and friction. The normal force is acting on the sphere and mg is acting in the downward direction. &#10;&#10; The forces which are acting parallel to the inclined plane are friction force, fs and the component of the weight of the cylinder,  mg \ sin &#952; . Normal Reaction is balanced by  mg\ cos&#952; . fs and  mg\ sin&#952;  are not in a line because they are R (Radius) distance apart. As a result, the torque will act, which gives rise to rotational motion i.e. rolling.

Question 13

A uniform sphere of mass  m  and radius  R  is placed on a rough horizontal surface (Figure). The sphere is struck horizontally at a height  h  from the floor. Match the following:

Accepted Answer

Mass of the sphere  = m Radius  = R h =   height of the floorThe sphere will roll without slipping when,  w \dfrac{v}{R} Where,  v =   linear velocity of the sphere w =  angular velocity of the sphere.Now, angular momentum of sphere, about centre of mass (applying conservation of angular momentum just before and after struch) mv (h - R) = I w = \left(\dfrac{2}{5} mR^2ight) \left(\dfrac{v}{R}ight) \Rightarrow mv (h - R) = \dfrac{2}{5} mvR \Rightarrow h - R = \dfrac{2}{5} R \Rightarrow h = \dfrac{7}{5} R Therefore, the sphere will roll without slipping with a constant velocity and hence, no loss of energy when  h = \dfrac{7}{5} R So, (D) matches (1)Now, Torque due to applied force about C.M., 	au = F (h - R)      [Clockwise]For  	au = 0, h = R  sphere will have only translational motion. It would lose energy by friction.Hence, (B) matches with (4).If  h > R, 	au ightarrow   positive, sphere will spin clockwise.So, (C) matches with (2)If  h <  R, 	au ightarrow   negative, sphere will spin anticlockwise.So, (A) matches with (3).

Question 14

A plank is moving with a velocity of  4\ m/sec . A disc of radius  1\ m  rolls without slipping on it with an angular velocity of  3\ rad/sec  as shown in figure. Find out the velocity of centre of the disc.

Accepted Answer

Since it is pure rolling the velocity of the point at the bottom will be equal to velocity of plank.So at  A { V }_{ cm }-RW=4\ { V }_{ cm }-3=4\ \Rightarrow { V }_{ cm }=7\quad m/s

Question 15

A disc of radius  R  rolls on a horizontal ground with linear acceleration  a  and angular acceleration  \alpha  as shown in figure. The magnitude of acceleration of point P at an instant when its linear velocity is  v  and angular velocity is  \omega  will be:

Accepted Answer

We have two components of acceleration of point P- one in radial direction which is  r \omega^2  and the other in tangential direction which is  r\alpha . Also, every point on the disc has a forward acceleration of a. Thus we have total acceleration in the forward direction as  a+r\alpha .Thus we get the resultant acceleration of point P as  \sqrt{(a+r\alpha)^2+(r\omega^2)^2} .

Question 16

A mass  m  hangs with the help of a string wrapped around a pulley on a frictionless bearing.The pulley has mass  m  and radius  R . Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass  m , if the string does not slip on the pulley, is :

Accepted Answer

Equations of motion are  mg-T = ma  ..... (i)   \displaystyle \mathrm{T}\cdot \mathrm{R}=\frac{1}{2}\mathrm{m}\mathrm{R}^{2} \alpha ..... (ii)  and  \mathrm{a}=\mathrm{R}\alpha  .... (iii) Solving this we get,  \displaystyle \mathrm{a}=\frac{2}{3}\mathrm{g} .

Question 17

A cylinder of mass  M  and radius  R  starts falling under gravity at  t=0  as shown in the figure. If the mass of chord is negligible, the tension in each string is?

Accepted Answer

At  t=ts, 7BD  will be as follows I_{cylinder}=\dfrac{MR^{2}}{2} The equation of translation and rotation motion will be as follows Mg-2T=Ma ..... (i) 2t.r=i_{cylinder}\alpha \Rightarrow 2T.R=\dfrac{MR^{2}}{2}\alpha ....... (ii) As the cylinder is rolling without sliping on the string a=R\alpha ..... (iii) From  (i), (ii)  and  (iii) , we get Mg-2T=4T \dfrac{Mg}{6}=T

Systems Of Particles And Rotational Motion