Rolling And Toppling (NEET LEVEL)

(i)           Velocity of centre of mass

(ii)          The time taken.

Solution

The point of contact has a forward velocity when the ball starts moving. This causes slipping and hence a kinetic friction equal to µmg acts in backward direction. This frictional force provides a clock-wise torque and hence the angular velocity increases from zero to ω in time t. The same frictional force reduces the linear velocity of centre of mass from v0 to v in time t.

If the ball begins to roll without slipping after time t,

                         v = rω

Using v = u + at for linear motion of centre of mass,

v = v0 – \displaystyle \left( {\frac{{\mu mg}}{m}} \right)\ t

v = v0 – μ gt                                                                    … (i)

Using ωf = ωi + at  andt = Ia for rotation:

w = 0 + \displaystyle \left( {\frac{{\mu mgr}}{I}} \right)\ t                                      … (ii)

We also have v = rω for pure rolling (iii)

t = \displaystyle \left( {\frac{{{{v}_{0}}{{k}^{2}}}}{{{{k}^{2}}+{{r}^{2}}}}} \right)\ \frac{1}{{\mu g}} (where I = mk2)

and  v = \displaystyle \frac{{{{v}_{0}}{{r}^{2}}}}{{{{k}^{2}}+{{r}^{2}}}}

For a solid sphere k2 = 2/5 r2

 time (t) =  \displaystyle \frac{{2{{v}_{0}}}}{{7\mu g}}  and   v = 5/7 v0

Hence after \displaystyle \frac{{2{{v}_{0}}}}{{7\mu g}} seconds, the translational speed of ball reduces to 5/7 v0 and it begins to roll without slipping.

Illustration

A solid cylinder of mass M and radius R rolls down an inclined plane with height h without slipping. Find the speed of its centre of mass when it reaches its bottom.

Solution

From COE, mgh  = \displaystyle \frac{1}{2}mv² + \displaystyle \frac{1}{2}Iω²

mgh  = \displaystyle \frac{1}{2}mv² + \displaystyle \frac{1}{2}. ( \displaystyle \frac{1}{2}mR²) ´ \displaystyle \frac{{{{v}^{2}}}}{{{{R}^{2}}}} = \displaystyle \frac{3}{4}mv²

v = \displaystyle \sqrt{{\frac{{4gh}}{3}}}

Illustration

A solid cylinder of radius 4 cm and mass 250 g rolls down an inclined plane (1 in 10). Find the acceleration and the total energy of the cylinder after 5 sec.

Solution:        Given that:R=4cm= 0.04m, M=250g = 0.25kg

\sin \alpha =\frac{1}{{10}},

The acceleration of a body rolling down an inclined plane is given by a=\frac{{g\sin \alpha }}{{1+\frac{{{{K}^{2}}}}{{{{r}^{2}}}}}}.

But the M.I. of the solid cylinder about its own axis =I=\frac{1}{2}M{{R}^{2}}=M{{K}^{2}}

Where K = radius of gyration

K=\frac{{{{K}^{2}}}}{{{{R}^{2}}}}=\frac{1}{2}

a=\frac{{g\sin \alpha }}{{1+\frac{1}{2}}} =\frac{2}{3}g\sin \alpha =\frac{2}{3}\times 9.8\times \frac{1}{{10}}\text{m/}{{\text{s}}^{2}}=0.653\ \text{m/}{{\text{s}}^{2}}

Now by 1^st equation of motion

v=u+at=0+0.653\times 5\ \text{m/s}=\text{3}\text{.26}\,\text{m/s}

 As the cylinder is rolling, the total K.E. of the cylinder is the sum of both K.E. of translation and rotation.

Total K.E. =\frac{1}{2}M{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}} =\frac{1}{2}M{{v}^{2}}+\frac{1}{2}\frac{1}{2}M{{R}^{2}}.\frac{{{{v}^{2}}}}{{{{R}^{2}}}}        \left( {\because \ \ \ \ \omega =\frac{v}{R}} \right)

=\frac{3}{4}M{{v}^{2}}=\frac{3}{4}\times 0.25\times {{(3.26)}^{2}}\text{J}=\text{2}\text{.0J}

Illustration

A ring, a disc and a sphere all of the same radius and mass roll down an inclined plane the same height h. Which of the three reaches the bottom (i) earliest, (ii) lastest?

Solution:        The acceleration of a body rolling down an inclined plane is given by,       

a=g\sin \alpha \left( {\frac{{M{{R}^{2}}}}{{I+M{{R}^{2}}}}} \right)

For a ring, I=M{{R}^{2}}

a=g\sin \alpha \left( {\frac{{M{{R}^{2}}}}{{M{{R}^{2}}+M{{R}^{2}}}}} \right)=\frac{1}{2}g\sin \alpha =0.5g\sin \alpha

For a disc, I=\frac{1}{2}M{{R}^{2}}

a=g\sin \alpha \left( {\frac{{M{{R}^{2}}}}{{\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}}}} \right)=\frac{2}{3}g\sin \alpha =0.67g\sin \alpha

For a sphere, I=\frac{2}{5}M{{R}^{2}}

a=g\sin \alpha \left( {\frac{{M{{R}^{2}}}}{{\frac{2}{5}M{{R}^{2}}+M{{R}^{2}}}}} \right)=\frac{5}{7}g\sin \alpha =0.71\ g\sin \alpha .