Question 727: Centre of Mass and Rotational Motion

Question

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant $$k$$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity $${\overrightarrow V _0} = {V_0}\widehat i$$. the coefficient of friction is $$\mu$$.

Accepted Answer

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.
$$	herefore -2kx+f=-M_{ac}$$ ..... (i)
Where $$a_c$$ = acceleration of centre of mass.
The torque acting on the disc about its centre of mass C is
$$	au  = f 	imes R = I 	imes {a_c}$$
$$	herefore$$ $$f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} 	imes {{{a_c}} \over R}$$
[$$	herefore$$ $$I = {1 \over 2}M{R^2},{a_c} = R{a_c}$$ for rolling without slipping]
$$	herefore$$ $$f = {1 \over 2}M{a_c}$$ ...... (ii)
From eq. (i) and (ii),
$$ \Rightarrow  - 2kx + f =  - 2f$$
$$ \Rightarrow f = {{2k} \over 3} 	imes x$$
Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $$x = A$$.
Where A = amplitude of S.H.M
Maximum frictional force,
$${f_{\max }} = {{2k} \over 3} 	imes A$$
The force should be almost equal to the limiting friction ($$\mu mg$$) for rolling without slipping.
$$	herefore$$ $$\mu mg = {{2k} \over 3} 	imes A$$ ..... (iv)
For S.H.M
Velocity amplitude $$ = A\omega $$
$$	herefore$$ $${V_0} = A\omega $$
$$	herefore$$ $${V_0} = {{3m\,mg} \over {2k}}w$$ (from (iv))
$$	herefore$$ $${V_0} = {{3\mu mg} \over {2k}} 	imes \sqrt {{{4k} \over {3M}}} $$ (As $$ \omega  = \sqrt {{{4k} \over {3M}}} $$)
$$	herefore$$ $${V_0} = mg\sqrt {{{3M} \over k}} $$

Answer

$$\mu g\sqrt {{M \over k}} $$

Answer

$$\mu g\sqrt {{M \over {2k}}} $$

Answer

$$\mu g\sqrt {{{3M} \over k}} $$

Answer

$$\mu g\sqrt {{{5M} \over {2k}}} $$

Centre of Mass and Rotational Motion

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