A solid cylinder having radius R and length L is slipping on a rough horizontal plane

Question

A solid cylinder having radius R and length L is slipping on a rough horizontal plane. At time $t = 0$ the cylinder has a translational velocity $v_{0} = 49$ m/s, perpendicular to its axis and a rotational velocity $v_{0} / 4 R$ about the centre. The time taken by the cylinder to start rolling is ______ seconds. (coefficient of kinetic friction $μ_{K}$ = 0.25 and g = 9.8 ${m/s}^{2}$ ) [2026]

Smart Achievers · Accepted Answer

(2)

$\vec{L} is conserved about A :$

$M v_{0} R + \frac{M R^{2}}{2} ω = M v R + \frac{M R^{2}}{2} ω$

$M v_{0} R + \frac{M R^{2}}{2} (\frac{v_{0}}{4 R}) = M v R + \frac{M R^{2}}{2} (\frac{v}{R})$

$M v_{0} R + \frac{M v_{0} R}{8} = M v R + \frac{M v R}{2}$

$\frac{9 M v_{0} R}{8} = \frac{3 M v R}{2} \Rightarrow v = \frac{3 v_{0}}{4}$

$v = v_{0} + a t$

$\frac{3 v_{0}}{4} = v_{0} - μ_{k} g t$

$μ_{k} g t = \frac{v_{0}}{4} \Rightarrow t = \frac{v_{0}}{4 μ_{k} g}$

$\Rightarrow t = \frac{v_{0}}{4 μ_{k} g} = \frac{49}{4 \times \frac{1}{4} \times 9.8} = 5 \sec$

Solution

1 15

2 5

3 10

4 7.5

Want Us to Email you About Special Offers & Updates?

Solution

1 15

2 5

3 10

4 7.5

Ans. (2) L→ is conserved about A: Mv0R+MR22ω=MvR+MR22ω Mv0R+MR22(v04R)=MvR+MR22(vR) Mv0R+Mv0R8=MvR+MvR2 9Mv0R8=3MvR2⇒v=3v04 v=v0+at 3v04=v0-μkgt μkgt=v04⇒t=v04μkg ⇒t=v04μkg=494×14×9.8=5 sec

Want Us to Email you About Special Offers & Updates?