Question 8: Centre of Mass and Rotational Motion

Question

A boy is pushing a ring of mass $$3 \mathrm{~kg}$$ and radius $$0.6 \mathrm{~m}$$ with a stick as shown in figure. The stick applies a force of $$3 \mathrm{~N}$$ 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

Accepted Answer

Here, mass of ring $$M=3 \mathrm{~kg}$$.
radius of ring $$r=0.6 \mathrm{~m}$$
force applied $$f=3 \mathrm{~N}$$
acceleration $$a=0.4 \mathrm{~m} / \mathrm{s}^2$$
coefficient of friction between stick and ring $$=\frac{F}{10}$$
Now, according figure $$f=M a$$
$$\begin{aligned}
3-f_s & =f=M a \
	ext{or} \quad f_s & =3-M a \
& =3-3 	imes 0.4=1.8 \mathrm{~N}
\end{aligned}$$
Now taking torque about $$O$$ We have
$$f_s R-f_k R=I \cdot \alpha$$
$$=M R^2 	imes \frac{a}{R}$$
$$\begin{array}{r}
\left(\because \alpha=	ext { angular acceleration }=\frac{a}{R}ight. \
	ext { and } \left.I=M R^2ight)
\end{array}$$
$$\begin{gathered}
	ext { or } 1.8 	imes 0.6-f_k 	imes 0.6=3 	imes 0.6 	imes 0.4 \
f_k=1.8-1.2=0.6 \mathrm{~N}
\end{gathered}$$
If $$\mu$$ is the coefficient of kinetic friction, then
$$\begin{aligned}
f_k & =\mu 	imes 3 \
\Rightarrow \quad \mu & =\frac{f_k}{2}=\frac{0.6}{3}=0.2=\frac{2}{10}=\frac{F}{10}
\end{aligned}$$
Hence on comparing, we get
$$F=2 \mathrm{~N}$$

Answer

2 N

Answer

4 N

Answer

6 N

Answer

3 N

Centre of Mass and Rotational Motion

Choose the correct answer: