Q: A solid sphere, a thin-walled hollow sphere, a solid cylinder, a thin-walled hollow cylinder and a ring, each of mass m and radius R , are simultaneously released at rest from top of an inclined plane, as shown in figure. The objects roll down the plane without slipping. Also we may consider the objects and the surface on which they roll to be perfectly rigid. Match column I and II

As seen from the attached concept, KE_{tot} is same for all the bodies.MI of ring and hollow cylinder are same for a given mass m and radius R.Hence, KE_{rot} is same for both ring and hollow cylinder.MI of I_{solid\: cylinder} \gt I_{hollow\: sphere} \gt I_{solid\: sphere} I_{ring} = I_{hollow\: cylinder} \gt I_{solid\: cylinder} \gt I_{hollow\: sphere} \gt I_{solid\: sphere} Hence, Rotational KE of the objects having the greater MI is more compared to others. \tau =I\alpha All objects are subjected to same torque. Object having least MI has the maximum angular acceleration.Hence, solid sphere has the max angular acceleration.Time taken to reach the bottom is maximum forRing or hollow cylinder.

Q: The pulleys in the figure are identical, each having a radius R and moment of inertia I . Find the acceleration of the block M .

Given in the question :- Assume, Acceleration =a angular acceleration, \propto =\cfrac { a }{ R } According to the question Mg-T_1 =Ma...(1) (T_2-T_1)R=Ia /R=\Rightarrow (T_2-T_1)=Ia /R^2 ....(2) (T_2-T_3)R=I/R^2 ....(3) \Rightarrow T_3 -mg=ma ....(4) By adding equation (2) and (3) we will get. \Rightarrow (T_1 -T_3)=2\ Ia/ R^2 ....(5) By adding equation (1) and (4) we will get. -mg+Mg+(T_3 -T_1)=Ma+ma Substituting the value for T_3 -T_1 we will get \Rightarrow Mg-mg=ma+ma+2Ia / R^2 \Rightarrow a=\dfrac {(M-m)g}{(M+m+2I/ R^2)}

Q: A heavy homogenous cylinder has a mass m and radius R. It is accelerated by a horizontal force F, which is applied through a rope wound around a light drum of radius r attached to the cylinder. The coefficient of static friction is sufficient for the cylinder to roll without slipping :Choose the correct options :

If the acceleration of the system is a , then- F - f = ma ...........(1)Now, Fr + fR = I\alpha Fr + fR = \dfrac{{m{R^2}}}{2}\alpha .Since there is no slipping, a = R\alpha Fr + fR = \dfrac{{mR}}{2}a \dfrac{{2Fr}}{R} + 2f = ma ..........(2)From equation 1 and 2 \dfrac{{2Fr}}{R} + 2f = F - f \dfrac{{2Fr}}{R} - F = - 3f f = \dfrac{F}{3}\left( {1 - \dfrac{{2r}}{R}} \right) Putting this in equation (i), we have- a = \dfrac{{2F}}{{3m}}\left( {\dfrac{{r + R}}{R}} \right) Now if, a > \dfrac{F}{m} \dfrac{{2F}}{{3m}}\left( {\dfrac{{r + R}}{R}} \right) > \dfrac{F}{m} \dfrac{2}{3}\left( {\dfrac{{r + R}}{R}} \right) > 1 2r + 2R > 3R 2r > R \dfrac{r}{R} > \dfrac{1}{2} Clearly, the direction of the friction will be forward.So, the correct options are (A) & (B)

Question 1

Find the radius of gyration of a hollow uniform sphere of radius R about its tangent.

Accepted Answer

Moment of inertia of a hollow sphere about a tangent,  I\,=\, \displaystyle\frac{5}{3}MR^2 MK^2\,=\, \displaystyle\frac{5}{3}MR^2\, \quad\, \Rightarrow\, K\, =\, \sqrt{\displaystyle\frac{5}{3}}R

Question 2

Identify the correct order in which the ratio of radius of gyration to radius increases for the following bodies.
I) rolling solid sphere
II) rolling solid cylinder
III) rolling hollow cylinder
IV) rolling hollow sphere

Accepted Answer

For solid sphere, the radius of gyration is  k=\sqrt{2R^2/5}=\sqrt{2/5}R For solid cylinder, the radius of gyration is  k=\sqrt{R^2/2}=\sqrt{1/2}R For hollow cylinder, the radius of gyration is  k=\sqrt{R^2}=R For hollow sphere, the radius of gyration is  k=\sqrt{2R^2/3}=\sqrt{2/3}R Arranging in increasing order,  I<II<IV<III

Question 3

A solid sphere, a thin-walled hollow sphere, a solid cylinder, a thin-walled hollow cylinder and a ring, each of mass

m

and radius

R

, are simultaneously released at rest from top of an inclined plane, as shown in figure. The objects roll down the plane without slipping. Also we may consider the objects and the surface on which they roll to be perfectly rigid. Match column I and II

Accepted Answer

As seen from the attached concept,  KE_{tot}  is same for all the bodies.MI of  ring and hollow cylinder are same for a given mass m and radius R.Hence,  KE_{rot}  is same for both ring and hollow cylinder.MI of  I_{solid\: cylinder} \gt I_{hollow\: sphere} \gt I_{solid\: sphere} I_{ring} = I_{hollow\: cylinder} \gt I_{solid\: cylinder} \gt I_{hollow\: sphere} \gt I_{solid\: sphere} Hence, Rotational KE of the objects having the greater MI is more compared to others. 	au =I\alpha All objects are subjected to same torque. Object having least MI has the maximum angular acceleration.Hence, solid sphere has the max angular acceleration.Time taken to reach the bottom is maximum forRing or hollow cylinder.

Question 4

The pulleys in the figure are identical, each having a radius  R  and moment of inertia  I . Find the acceleration of the block  M .

Accepted Answer

Given in the question  :- Assume, Acceleration =a angular acceleration,  \propto =\cfrac { a }{ R }  According to the question Mg-T_1 =Ma...(1) (T_2-T_1)R=Ia /R=\Rightarrow (T_2-T_1)=Ia /R^2 ....(2) (T_2-T_3)R=I/R^2  ....(3) \Rightarrow T_3 -mg=ma  ....(4) By adding equation  (2)  and  (3)  we will get. \Rightarrow (T_1 -T_3)=2\ Ia/ R^2  ....(5) By adding equation  (1)  and  (4)  we will get. -mg+Mg+(T_3 -T_1)=Ma+ma Substituting the value for  T_3 -T_1  we will get \Rightarrow Mg-mg=ma+ma+2Ia / R^2 \Rightarrow a=\dfrac {(M-m)g}{(M+m+2I/ R^2)}

Question 5

Question 6

A heavy homogenous cylinder has a mass m and radius R. It is accelerated by a horizontal force F, which is applied through a rope wound around a light drum of radius r attached to the cylinder. The coefficient of static friction is sufficient for the cylinder to roll without slipping :
Choose the correct options :

Accepted Answer

If the acceleration of the system is  a , then- F - f = ma  ...........(1)Now, Fr + fR = I\alpha  Fr + fR = \dfrac{{m{R^2}}}{2}\alpha  .Since there is no slipping,  a = R\alpha  Fr + fR = \dfrac{{mR}}{2}a \dfrac{{2Fr}}{R} + 2f = ma   ..........(2)From equation 1 and 2                                         \dfrac{{2Fr}}{R} + 2f = F - f \dfrac{{2Fr}}{R} - F =  - 3f f = \dfrac{F}{3}\left( {1 - \dfrac{{2r}}{R}} \right) Putting this in equation (i), we have- a = \dfrac{{2F}}{{3m}}\left( {\dfrac{{r + R}}{R}} \right) Now if, a > \dfrac{F}{m} \dfrac{{2F}}{{3m}}\left( {\dfrac{{r + R}}{R}} \right) > \dfrac{F}{m} \dfrac{2}{3}\left( {\dfrac{{r + R}}{R}} \right) > 1 2r + 2R > 3R 2r > R \dfrac{r}{R} > \dfrac{1}{2} Clearly, the direction of the friction will be forward.So, the correct options are (A) & (B)

Systems Of Particles And Rotational Motion