Question

A wheel (disc of mass $m$, radius $R$) is mounted on a base of mass $M$. The whole system is placed on rough horizontal surface. The coefficient of friction between the base and the surface is $\mu$. Initially the system is at rest. A time varying horizontal force $F = 2t$ is applied tangentially on the wheel, where $t$ is time in second and force $F$ is in newton. The kinetic energy of the disc when the system is about to slide on ground is :-

• A. $\frac{1}{m} (\frac{\mu (M) g}{2})^4$
• B. $\frac{1}{m} (\frac{\mu (m+M) g}{2})^4$
• C. $\frac{1}{m} (\frac{\mu (m+M) g}{2})^2$
• D. $\frac{1}{m} (\frac{\mu (M) g}{2})^2$

Answer 1

Answer: (B)

$\frac{1}{m} (\frac{\mu (m+M) g}{2})^4$

Answer 2

The problem describes a system consisting of a wheel (disc) of mass $m$ and radius $R$, mounted on a base of mass $M$. The entire system is placed on a rough horizontal surface with a coefficient of friction $\mu$. A time-varying horizontal force $F = 2t$ is applied tangentially to the top of the wheel. The axle of the wheel is smooth, meaning there is no friction between the wheel and the base at the pivot point. We need to find the kinetic energy of the disc when the system is about to slide on the ground.

1. Identify the forces acting on the system (wheel + base). The total downward force is $(m+M)g$.
2. The normal force from the ground is $N_G = (m+M)g$.
3. The maximum static friction force the ground can exert on the base is $f_{s,max} = \mu N_G = \mu (m+M)g$.
4. The system is "about to slide" when the applied horizontal force $F$ equals $f_{s,max}$.
5. Given $F = 2t$, we set $2t_s = \mu (m+M)g$ to find the time $t_s$ at which sliding is imminent. $t_s = \frac{\mu (m+M)g}{2}$.
6. Up to this point, the system (base + wheel's center of mass) remains at rest, as the static friction balances the applied force. Therefore, the linear velocity of the disc's center of mass is $v=0$ at $t_s$.
7. The kinetic energy of the disc is thus purely rotational: $KE_{disc} = \frac{1}{2} I \omega^2$. For a disc, $I = \frac{1}{2} m R^2$.
8. The applied force $F$ causes the wheel to rotate. The torque on the wheel is $\tau = F R$. $\tau = I \alpha \implies F R = \frac{1}{2} m R^2 \alpha$.
9. Substitute $F=2t$: $2t R = \frac{1}{2} m R^2 \alpha \implies \alpha = \frac{4t}{mR}$.
10. The angular velocity $\omega$ is the integral of angular acceleration $\alpha$ from $t=0$ to $t_s$. $\omega(t_s) = \int_0^{t_s} \alpha(t') dt' = \int_0^{t_s} \frac{4t'}{mR} dt' = \frac{4}{mR} \left[ \frac{(t')^2}{2} \right]_0^{t_s} = \frac{2t_s^2}{mR}$.
11. Substitute the expression for $t_s$: $\omega(t_s) = \frac{2}{mR} \left( \frac{\mu (m+M)g}{2} \right)^2 = \frac{2}{mR} \frac{\mu^2 (m+M)^2 g^2}{4} = \frac{\mu^2 (m+M)^2 g^2}{2mR}$.
12. Calculate the kinetic energy of the disc: $KE_{disc} = \frac{1}{2} \left( \frac{1}{2} m R^2 \right) \omega(t_s)^2 = \frac{1}{4} m R^2 \left( \frac{\mu^2 (m+M)^2 g^2}{2mR} \right)^2$ $KE_{disc} = \frac{1}{4} m R^2 \frac{\mu^4 (m+M)^4 g^4}{4m^2 R^2} = \frac{\mu^4 (m+M)^4 g^4}{16m}$.
13. This can be written as $KE_{disc} = \frac{1}{m} \left( \frac{\mu (m+M) g}{2} \right)^4$.