Question

A boy is pushing a ring of mass $3\, kg$ and radius $0.6\, m$ with a stick as shown in figure. The stick applies a force of $3\, N$ on the ring and rolls it without slipping with an acceleration of $0.4\, m / s ^{2}$. 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 $M=3\, kg$.
radius of ring $r=0.6\, m$
force applied $f=3\, N$
acceleration $a=0.4\, m / s ^{2}$
coefficient of friction between
stick and ring $=\frac{F}{10}$
Now, according figure $f=M a$
$3-f_{s} =f=M a$
or $f_{s}=3-M a$
$=3-3 \times 0.4=1.8\, N$
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 $1.8 \times 0.6-f_{k} \times 0.6=3 \times 0.6 \times 0.4$
$f_{k}=1.8-1.2=0.6\, N$
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
$F=2\, N$