Question

A solid sphere of mass 0.5 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is $\dfrac{2}{7}$. The maximum force that can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface is
A) 2.3 N
B) 3.3 N
C) 6 N
D) None of these.

Answer 1

As the body is a solid sphere and as it is not slipping i.e. it is rolling than the applied force and the frictional force is acting in the same direction. These forces will give us some torque. Equating these we can calculate the maximum for that can be applied without slipping.

Complete step by step answer:
Here, two main forces are active on the solid sphere. One is the applied force and the second is the frictional force. Now if both these force in the opposite direction then it will slip, but if they are in the same direction then the sphere will not slip and it will just start rolling.

Thus, we have the force applied ${F_a}$ and frictional force${F_{fric}}$ and let both of them be applied towards + x-direction.
Now, Torque produced is given by,
$\tau = I\alpha$,
where I= moment of inertia and $\alpha$is the translational acceleration.
Again, torque can be written as,
$\tau = {F_a}R - {F_{fric}}R$
Now, from the above equation, we have,
$I\alpha = {F_a} + {F_{fric}}$
Thus, rewriting the equation we get,
$\Rightarrow \alpha = \dfrac{{\left( {{F_a} - {F_{fric}}} \right)R}}{I}$
Now, putting the value of I,
$\Rightarrow \alpha = \dfrac{{\left( {{F_a} - {F_{fric}}} \right)R}}{{2m{R^2}}} \times 5$
Replacing the frictional force in the above equation,
$\Rightarrow \alpha = \dfrac{{\left( {{F_a} - \mu mg} \right)R}}{{2m{R^2}}} \times 5$
Simplifying we get,
$\Rightarrow \alpha = \dfrac{5}{2}\dfrac{{\left( {{F_a} - \mu mg} \right)}}{{mR}}$
Also, acceleration is,
$\Rightarrow a = \dfrac{{{F_a} + {F_{fric}}}}{m} \\\ \Rightarrow a = \dfrac{{{F_a} + \mu mg}}{m} \\\$
Now, as the sphere is not slipping, we have,
$a = \alpha R$

Replacing the values, we get,
$\Rightarrow \dfrac{{{F_a} + \mu mg}}{m} = \dfrac{5}{2}\dfrac{{\left( {{F_a} - \mu mg} \right)}}{{mR}}R \\\ \Rightarrow {F_a} + \mu mg = \dfrac{5}{2}\left( {{F_a} - \mu mg} \right) \\\ \Rightarrow 2{F_a} + 2\mu mg = 5{F_a} - 5\mu mg \\\ \Rightarrow 5{F_a} - 2{F_a} = 5\mu mg + 2\mu mg \\\ \Rightarrow 3{F_a} = 7\mu mg \\\ \Rightarrow {F_a} = \dfrac{7}{3}\mu mg \\\$
Now, we have maximum force as,
$\Rightarrow {F_{{a_{\max }}}} = \dfrac{7}{3}\mu mg \\\ \Rightarrow {F_{{a_{\max }}}} = \dfrac{7}{3} \times \dfrac{2}{7}mg \\\ \Rightarrow {F_{{a_{\max }}}} = \dfrac{2}{3}mg \\\$
Again, putting the value m= 0.5 kg and g=10 $m{s^{ - 1}}$ we have,
$\Rightarrow {F_{{a_{\max }}}} = \dfrac{2}{3}mg \\\ \Rightarrow {F_{{a_{\max }}}} = \dfrac{2}{3} \times \dfrac{1}{2} \times 10 \\\ \Rightarrow {F_{{a_{\max }}}} = \dfrac{{10}}{3} \\\ \Rightarrow {F_{{a_{\max }}}} = 3.3 \\\$

Thus, the maximum force that can be applied for the sphere to roll but not slip is $3.3 N$. So, option (B) is the right answer.

Note:
Here, we have to specially make note of the sphere rolling condition. This condition will allow us to equate to get the proper output. Most students miss out on this particular point and are unable to solve a problem of this kind.