MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Problem Solving

In the arrang...

Question

In the arrangement shown in the figure, the mass of the uniform solid cylinder of radius $R$ is equal to $m$ and the masses of two bodies are equal to $m_{1}$ and $m_{2}$ . The thread slipping and the friction in the axle of the cylinder are supposed to the absent. Find the angular acceleration of the cylinder and the ratio of the tensions $\frac{T_{1}}{T_{2}}$ of the sections of the thread in the process of motion.

Open in App

Solution

Because the pulley has got a mass hence tension in the two strings will be different, Hence equation of motion for mass

$m_{1} g - T = m_{1} a$

For mass $m_{2}; T_{2} - m_{2} g = m_{2} a ⟶ (i i)$

For rotational motion of cylinder,

$(T - T_{2}) R = I α ⟶ (i i i) α ⟶$ angular acceleration

Adding equations $(i)$ and $(i i)$

$m_{1} g - T + T_{2} - m_{2} g = (m_{1} + m_{2}) a (m_{1} - m_{2}) g - (T - T_{2}) = (m_{1} + m_{2}) α R (m_{1} - m_{2}) g - (I α / R) = (m_{1} + m_{2}) α R (m_{1} - m_{2}) g = [m_{1} + m_{2} + (I / R^{2})] α R ∴ α = \frac{(m_{1} - m_{2}) g}{[R (m_{1} + m_{2} + I / R^{2})]} = \frac{2 (m_{1} - m_{2}) g}{R (m_{1} + m_{2} + \frac{1}{2} (\frac{m R^{2}}{R^{2}}))}$

$= \frac{2 (m_{1} - m_{2}) g}{R (2 m_{1} + {2 m}_{2} + m)}$

Form equation $(i i)$ and $(i i i)$ we get,

$\frac{T_{1}}{T_{2}} - 1 = \frac{I α}{R T_{2}} = \frac{I α}{R (m_{2} a + m_{2} g)} = \frac{I α}{R^{2} m_{2} [α + (\frac{g}{12})]} = \frac{I}{R^{2} m_{2} [1 + (\frac{g}{12})]} = \frac{I}{R^{2} m_{2} [1 + {g ((2 m_{1} + {2 m}_{2} + m) R / 2 (m_{1} - m_{2}) g R)}]}$

$\frac{T_{1}}{T_{2}} - 1 = \frac{\frac{1}{2} m R^{2} \times 2 (m_{1} - m_{2})}{R^{2} m_{2} [2 m_{1} - 2 m_{2} + 2 m_{1} + 2 m_{2} + m]} = \frac{m (m_{1} - m_{2})}{m_{2} (4 m_{1} + m)} ∴ \frac{T_{1}}{T_{2}} = \frac{m (m_{1} - m_{2})}{m_{2} (4 m_{1} + m)} + 1 \frac{T_{1}}{T_{2}} = \frac{m_{1} (4 m_{2} + m)}{m_{2} (4 m_{1} + m)}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In the arrangement shown in the figure the mass of the uniform solid cylinder of Radius

R

m

and the masses of two bodies are equal to

m_{1}

m_{2}

. The thread slipping and the friction in the axel of the cylinder are supposed to be absent. Find the angular acceleration of the cylinder and the ratio of tensions

\frac{T_{1}}{T_{2}}

of the verticle sections at the thread in the process of motion.

Q. The arrangement shown in figure consists of two identical, uniform, solid cylinders, each of mass

m

, on which two light threads are wound symmetrically.
Find the tensions of each thread in the process of motion. The friction in the axle of the upper cylinder is assumed to be absent.

Q. In the arrangement of figure shown above, the masses

m_{0}

m_{1}

m_{2}

of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration

\to w

with which the body

m_{0}

comes down, and the tension of the thread binding together the bodies

m_{1}

m_{2}

, if the coefficient of friction between these bodies and the horizontal surface is equal to

k

. Consider possible cases.

Q. In the arrangement shown in figure above the bodies have masses

m_{0}

m_{1}

m_{2}

, the friction is absent, the masses of the pulleys and the threads are negligible. Find the acceleration of the body

m_{1}

. Look into possible cases.

Q. A string is wrapped on the surface of a solid cylinder as shown in the figure. If the cylinder is released, then find the acceleration of the centre of mass of the cylinder and the tension in the string respectively. Mass of the cylinder is

M

r

. (No slipping occurs between any contact points)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Problem Solving Using Newton's Laws

PHYSICS

Watch in App

explore_more_icon

Explore more

Problem Solving

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon