Question-216665

Question Number 216665 by Tawa11 last updated on 14/Feb/25

Commented by Tawa11 last updated on 14/Feb/25

In the figure, a cylinder of mass M, radius R
and moment of inertia I = (MR²)/2 is placed
on a horizontal table. The rope wound around
the cylinder passes over an ideal massless
pulley and is then tied to a block of mass,
m = 2M.

At the instant shown, the block is falling and
the cylinder rolls without slipping, unwinding
the rope. Find:
a) the acceleration of the block,
b) the acceleration of the center of mass of
the spool

Answered by mr W last updated on 17/Feb/25

Commented by mr W last updated on 16/Feb/25

mA=mg−T ⇒T=m(g−A)=2M(g−2a) (((MR^2 )/2)+MR^2 )(a/R)=2RT ⇒T=((3Ma)/4) ((3Ma)/4)=2M(g−2a) 19a=8g ⇒a=((8g)/(19)) ✓ ⇒A=((32g)/(19)) ✓

$${mA}={mg}−{T} \\ $$$$\Rightarrow{T}={m}\left({g}−{A}\right)=\mathrm{2}{M}\left({g}−\mathrm{2}{a}\right) \\ $$$$\left(\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}+{MR}^{\mathrm{2}} \right)\frac{{a}}{{R}}=\mathrm{2}{RT} \\ $$$$\Rightarrow{T}=\frac{\mathrm{3}{Ma}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}{Ma}}{\mathrm{4}}=\mathrm{2}{M}\left({g}−\mathrm{2}{a}\right) \\ $$$$\mathrm{19}{a}=\mathrm{8}{g} \\ $$$$\Rightarrow{a}=\frac{\mathrm{8}{g}}{\mathrm{19}}\:\checkmark \\ $$$$\Rightarrow{A}=\frac{\mathrm{32}{g}}{\mathrm{19}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 15/Feb/25