Question

An annular disc of mass M, inner radius a and outer radius b is placed on a horizontal surface with a coefficient of friction μ, as shown in the figure. At some time, an impulse J0 $\hat{x}$ is applied at a height of h at the center of the disc. If h=hm, then the disc rolls without slipping along the x-axis, which of the following statement is/are correct?

• A. For h=hm the initial angular velocity doesn't depend on the inner radius a.
• B. For μ≠0 & a$\rightarrow$b, hm=b
• C. For μ=0 & h=0 the wheel always slides without rolling
• D. For μ≠0 & a$\rightarrow$0, hm=$\frac{b}{2}$

Answer 1

Answer: (A), (B), (C), (D)

For h=hm the initial angular velocity doesn't depend on the inner radius a.

Answer 2

Angular impulse about $CM=\Delta L$
$J_0\,h=I_{cm}\omega$
$\omega=\frac{J_0h}{I_{cm}}$
Linear impulse = $\Delta P$
$J_0=\Delta P$
$J_0=MV-0$
$V=\frac{J_0}{M}$
As disc is pure rolling,
$V=\omega b$
$\omega=\frac{V}{b}$
$\omega=\frac{J_0}{Mb}$
$\omega$ is independent of radius a…… (Ans.1)
Now, If $a\rightarrow b$, then the annular disc is a ring of radius b
$I_{cm}=Mb^2$
$\omega=\frac{J_0h}{Mb^2}$
$\frac{J_0}{Mb}=\frac{J_0h}{Mb^2}$
$h=b$
$\therefore h_m=b$…….(Ans.2)

if h=0 and $\mu$=0,
as h=0 $\,\,\,\,\omega=\frac{J_0h}{Mb^2}$ = 0
but $V=\frac{J_0}{M}$………(Ans.3)
If a$\rightarrow$0then annular disc is disc of radius b
$I_{cm}=\frac{Mb^2}{2}$
$\omega=\frac{J_0h\times2}{Mb^2};h=\frac{b}{2};h_m=\frac{b}{2}$……..(Ans.4)