A tangential force F acts at the top of a disc of mass m and radius R. If it rolls without slipping. Then

Question

A tangential force F acts at the top of a disc of mass m and radius R. If it rolls without slipping. Then (A) Acceleration of disc = (2F/3m) (B) Friction force between disc and surface= (2F/3) (C) Acceleration of disc= (6F/5m) (D) Friction force between disc and surface is = (F/3)

Tardigrade Admin · Accepted Answer

Equation regarding translatory motion F + f = ma ...(i) Equation regarding rotational motion FR - fR = (mR2/2) (a/R) ...(ii) From equation (i) and (ii) 2F= (3ma/2) or a = (4F/3m) Also, F+f = (m4F/3m) f = (4/3)F -F = (F/3)

Q. A tangential force $F$ acts at the top of a disc of mass $m$ and radius $R$ . If it rolls without slipping. Then

Acceleration of disc $= \frac{2 F}{3 m}$

Friction force between disc and surface $= \frac{2 F}{3}$

Acceleration of disc $= \frac{6 F}{5 m}$

Friction force between disc and surface is $= \frac{F}{3}$

Equation regarding translatory motion
$F + f = ma ... (i)$
Equation regarding rotational motion
$FR - f R = \frac{m R ^{2}}{2} \frac{a}{R} ... (ii)$
From equation $(i)$ and $(ii)$
$2 F = \frac{3 ma}{2}$ or $a = \frac{4 F}{3 m}$
Also, $F + f = \frac{m 4 F}{3 m}$
$f = \frac{4}{3} F - F = \frac{F}{3}$

Q. A tangential force F acts at the top of a disc of mass m and radius R. If it rolls without slipping. Then

Acceleration of disc =3m2F​

Friction force between disc and surface=32F​

Acceleration of disc=5m6F​

Friction force between disc and surface is =3F​