Question

A boy is pushing a ring of mass $3kg$ and radius $0.6m$ with a stick as shown in figure. The stick applies a force of $3N$ on the ring and rolls it without slipping with an acceleration of $0.4m/s^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $\frac{F}{10}$. The value of $F$ is

• A. 2 N
• B. 4 N
• C. 6 N
• D. 3 N

Answer 1

Answer: (A)

Here, mass of ring $M=3kg$.
radius of ring $r=0.6m$
force applied $f=3N$
acceleration $a=0.4m/s^2$
coefficient of friction between
stick and ring $=\frac{F}{10}$
Now, according figure $f=Ma$
$3-f_s=f=Ma$
or $f_s=3-Ma$
$=3-3\times 0.4=1.8N$
Now taking torque about $O$
We have $f_{s}R-f_{k}R=I\cdot \alpha$

$=M{R}^2\times \frac{a}{R}$
($\because \alpha =$ angular acceleration $=\frac{a}{R}$
and $I=M{R}^2$)
or $1.8\times 0.6-f_k\times 0.6=3\times 0.6\times 0.4$
$f_k=1.8-1.2=0.6N$
If $\mu$ is the coefficient of kinetic friction, then
$f_k=\mu \times 3$
$\mu =\frac{f_k}{2}=\frac{0.6}{3}=0.2$
$=\frac{2}{10}=\frac{F}{10}$
Hence on compairing, we get
$F=2N$