A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc -

Question

A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc - (A) velocity is v, acceleration is zero (B) velocity is zero, acceleration is zero (C) velocity is v, acceleration is v2/R (D) velocity is zero, acceleration is v2/R

Tardigrade Admin · Accepted Answer

v A =0 (due to pure rolling) Angular acceleration (for A), a =R w2=R (v2/R2) ⇒ a=(v2/R)

Q. A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc -

velocity is v, acceleration is zero

velocity is zero, acceleration is zero

velocity is v, acceleration is $v^{2} / R$

velocity is zero, acceleration is $v^{2} / R$

$v_{A} = 0$ (due to pure rolling)
Angular acceleration (for A), a
$= R w^{2} = R \frac{v ^{2}}{R ^{2}}$
$\Rightarrow a = \frac{v ^{2}}{R}$

Q. A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc -

velocity is v, acceleration is zero

velocity is zero, acceleration is zero

velocity is v, acceleration is v2/R

velocity is zero, acceleration is v2/R

vA​=0 (due to pure rolling) Angular acceleration (for A), a =Rw2=RR2v2​ ⇒a=Rv2​

velocity is v, acceleration is $v^{2} / R$

velocity is zero, acceleration is $v^{2} / R$

$v_{A} = 0$ (due to pure rolling)
Angular acceleration (for A), a
$= R w^{2} = R \frac{v ^{2}}{R ^{2}}$
$\Rightarrow a = \frac{v ^{2}}{R}$