
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:
\[\begin{align}
& A.\dfrac{2g}{3}and\dfrac{mg}{3} \\
& B.gand\dfrac{mg}{2} \\
& C.\dfrac{g}{3}and\dfrac{mg}{2} \\
& A.\dfrac{g}{2}and\dfrac{mg}{3} \\
\end{align}\]
Answer
522.9k+ views
Hint: 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: $\tau=f\times d$, $\tau=I\times \omega$ and $I=\dfrac{1}{2}mr^{2}$
Complete answer:
Let the mass of the given cylinder which is wound up by the strings be $m$, and $r$ its radius.
Then the moment of inertia experienced by the cylinder during a rotation is given by $I=\dfrac{1}{2}mr^{2}$
Then the torque experienced by the cylinder is given as $\tau=I\times \omega$ where $\omega$ is the angular acceleration. We know that the angular acceleration is given as $\omega=\dfrac{a}{r}$ where $a$ is the linear acceleration and $r$ is the radius of the cylinder.
Also the torque is given as $\tau=f\times d$ where $f$ is the force acting on the cylinder at a perpendicular distance $d$. Here $d=r$ and $f=T$ where $T$ is the tension on the strings which are wound around the cylinder.
Substituting the values we get, $T\times r=\dfrac{1}{2}mr^{2}\times \dfrac{a}{r}$
Or, $T=\dfrac{1}{2}ma$
Also $F_{net}=W-T$ where $W=mg$ is the weight of the cylinder.
Then, $F_{net}=mg-\dfrac{1}{2}ma$
We know that the net force is given as $F_{net}=ma$ .
Then, $ma=mg-\dfrac{1}{2}ma$
$\Rightarrow mg=\dfrac{3}{2}ma$
$\Rightarrow a=\dfrac{2g}{3}$
Substituting the value of $a$ in $T$ we get, $T=\dfrac{1}{2}ma=\dfrac{mg}{3}$.
Hence we get $a=\dfrac{2g}{3}$ and $T=\dfrac{mg}{3}$
Thus the answer is \[A.\dfrac{2g}{3}and\dfrac{mg}{3}\].
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.
Formula: $\tau=f\times d$, $\tau=I\times \omega$ and $I=\dfrac{1}{2}mr^{2}$
Complete answer:
Let the mass of the given cylinder which is wound up by the strings be $m$, and $r$ its radius.
Then the moment of inertia experienced by the cylinder during a rotation is given by $I=\dfrac{1}{2}mr^{2}$
Then the torque experienced by the cylinder is given as $\tau=I\times \omega$ where $\omega$ is the angular acceleration. We know that the angular acceleration is given as $\omega=\dfrac{a}{r}$ where $a$ is the linear acceleration and $r$ is the radius of the cylinder.
Also the torque is given as $\tau=f\times d$ where $f$ is the force acting on the cylinder at a perpendicular distance $d$. Here $d=r$ and $f=T$ where $T$ is the tension on the strings which are wound around the cylinder.
Substituting the values we get, $T\times r=\dfrac{1}{2}mr^{2}\times \dfrac{a}{r}$
Or, $T=\dfrac{1}{2}ma$
Also $F_{net}=W-T$ where $W=mg$ is the weight of the cylinder.
Then, $F_{net}=mg-\dfrac{1}{2}ma$
We know that the net force is given as $F_{net}=ma$ .
Then, $ma=mg-\dfrac{1}{2}ma$
$\Rightarrow mg=\dfrac{3}{2}ma$
$\Rightarrow a=\dfrac{2g}{3}$
Substituting the value of $a$ in $T$ we get, $T=\dfrac{1}{2}ma=\dfrac{mg}{3}$.
Hence we get $a=\dfrac{2g}{3}$ and $T=\dfrac{mg}{3}$
Thus the answer is \[A.\dfrac{2g}{3}and\dfrac{mg}{3}\].
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
