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Question: A string is wrapped several times round a solid cylinder and then the end of the string is held stat...

A string is wrapped several times round a solid cylinder and then the end of the string is held stationary while the cylinder is released from rest with an initial motion. The acceleration of the cylinder and tension in the string will be:

& A.\dfrac{2g}{3}and\dfrac{mg}{3} \\\ & B.gand\dfrac{mg}{2} \\\ & C.\dfrac{g}{3}and\dfrac{mg}{2} \\\ & A.\dfrac{g}{2}and\dfrac{mg}{3} \\\ \end{aligned}$$
Explanation

Solution

To find the tension is the string, we need to calculate the net force of the string. Since the cylinder undergoes rotation when the string is pulled, we can find its torque which in turn is used to calculate the acceleration of the cylinder.
Formula: τ=f×d\tau=f\times d, τ=I×ω\tau=I\times \omega and I=12mr2I=\dfrac{1}{2}mr^{2}

Complete answer:
Let the mass of the given cylinder which is wound up by the strings be mm, and rr its radius.
Then the moment of inertia experienced by the cylinder during a rotation is given by I=12mr2I=\dfrac{1}{2}mr^{2}
Then the torque experienced by the cylinder is given as τ=I×ω\tau=I\times \omega where ω\omega is the angular acceleration. We know that the angular acceleration is given as ω=ar\omega=\dfrac{a}{r} where aa is the linear acceleration and rr is the radius of the cylinder.
Also the torque is given as τ=f×d\tau=f\times d where ff is the force acting on the cylinder at a perpendicular distance dd. Here d=rd=r and f=Tf=T where TT is the tension on the strings which are wound around the cylinder.
Substituting the values we get, T×r=12mr2×arT\times r=\dfrac{1}{2}mr^{2}\times \dfrac{a}{r}
Or, T=12maT=\dfrac{1}{2}ma
Also Fnet=WTF_{net}=W-T where W=mgW=mg is the weight of the cylinder.
Then, Fnet=mg12maF_{net}=mg-\dfrac{1}{2}ma
We know that the net force is given as Fnet=maF_{net}=ma .
Then, ma=mg12mama=mg-\dfrac{1}{2}ma
mg=32ma\Rightarrow mg=\dfrac{3}{2}ma
a=2g3\Rightarrow a=\dfrac{2g}{3}
Substituting the value of aa in TT we get, T=12ma=mg3T=\dfrac{1}{2}ma=\dfrac{mg}{3}.
Hence we get a=2g3a=\dfrac{2g}{3} and T=mg3T=\dfrac{mg}{3}

Thus the answer is A.2g3andmg3A.\dfrac{2g}{3}and\dfrac{mg}{3}.

Note:
Here, we are assuming the wound up strings as a solid cylinder with some mass and radius. This is used to simplify the problem and helps in the easy visualisation of the problem. Here, we consider the angular acceleration of the cylinder in terms of the linear acceleration as the string undergoes linear acceleration.