JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 7)
A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :
_8th_April_Morning_Slot_en_7_1.png)
$${{2\,\,mg} \over {2\,m + M}}$$
$${{2\,\,Mg} \over {2\,m + M}}$$
$${{2\,\,mg} \over {2\,M + m}}$$
$${{2\,\,Mg} \over {2\,M + M}}$$
Explanation
_8th_April_Morning_Slot_en_7_2.png)
From figure, we can say
ma = mg $$-$$ T . . . . . (1)
Moment of inertia of a uniform disc,
I = $${1 \over 2}$$ MR2
and the acceleration of the disc, a = $$ \propto $$ R
We know,
Torque = I$$ \propto $$ = RT
$$\therefore\,\,\,$$ RT = $${1 \over 2}$$ M R2 $$ \times $$ $${a \over R}$$ = $${{MaR} \over 2}$$
$$ \Rightarrow $$ $$\,\,\,$$ T = $${{Ma} \over 2}$$
Putting the value of T in eq (1).
ma = mg $$-$$ $${{Ma} \over 2}$$
$$ \Rightarrow $$ $$\,\,\,$$ a ( m + $${{Ma} \over 2}$$) = mg
$$ \Rightarrow $$ $$\,\,\,$$ a = $${{2mg} \over {2m + M}}$$
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