A reel of thread unrolls itself falling under gravity. Neglecting mass of the thread, the acceleration of the reel is

Question

A reel of thread unrolls itself falling under gravity. Neglecting mass of the thread, the acceleration of the reel is (A)  g  (B)  (g/2)  (C)  (2g/3)  (D)  (4g/3)

Tardigrade Admin · Accepted Answer

a=(g/(1+(k2)R2)) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/48601546e05fe666968e312e9d94df76-.png" /> For solid cylinder (K2/R2)=(1/2) ∴ a=(g/1+(1)2)=(2/3) g

Q. A reel of thread unrolls itself falling under gravity. Neglecting mass of the thread, the acceleration of the reel is

$g$

$\frac{g}{2}$

$\frac{2 g}{3}$

$\frac{4 g}{3}$

$a = \frac{g}{( 1 + \frac{k ^{2}}{R ^{2}} )}$

For solid cylinder $\frac{K ^{2}}{R ^{2}} = \frac{1}{2}$
$∴ a = \frac{g}{1 + \frac{1}{2}} = \frac{2}{3} g$

Q. A reel of thread unrolls itself falling under gravity. Neglecting mass of the thread, the acceleration of the reel is

g

2g​

32g​

34g​

a=(1+R2k2​)g​ For solid cylinder R2K2​=21​ ∴a=1+21​g​=32​g

$g$

$\frac{g}{2}$

$\frac{2 g}{3}$

$\frac{4 g}{3}$

$a = \frac{g}{( 1 + \frac{k ^{2}}{R ^{2}} )}$

For solid cylinder $\frac{K ^{2}}{R ^{2}} = \frac{1}{2}$
$∴ a = \frac{g}{1 + \frac{1}{2}} = \frac{2}{3} g$