SATHEE: Rotation 5 Question 29

Rotation 5 Question 29

41. A block $X$ of mass $0.5 k g$ is held by a long massless string on a frictionless inclined plane of inclination $30^{\circ}$ to the horizontal. The string is wound on a uniform solid cylindrical drum $Y$ of mass $2 k g$ and of radius $0.2 m$ as shown in figure.

The drum is given an initial angular velocity such that the block $X$ starts moving up the plane.

$(1994, 6$ M)

(a) Find the tension in the string during the motion.

(b) At a certain instant of time, the magnitude of the angular velocity of $Y$ is $10 r a d s^{- 1}$ . Calculate the distance travelled by $X$ from that instant of time until it comes to rest.

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Answer:

Correct Answer: 41. (a) $1.63 N$

(b) $1.22 m$

Solution:

Given, mass of block $X, m = 0.5 k g$

Mass of drum $Y, M = 2 k g$

Radius of drum, $R = 0.2 m$

Angle of inclined plane, $θ = 30^{\circ}$

(a) Let $a$ be the linear retardation of block $X$ and $α$ be the angular retardation of $drum Y$ . Then, $a = R α$

$m g \sin 30^{\circ} - T = m a$

$\begin{aligned} \frac{m g}{2} - T & = m a \\ α = \frac{τ}{I} & = \frac{T R}{\frac{1}{2} M R^{2}} \\ α & = \frac{2 T}{M R} \end{aligned}$

Solving Eqs. (i), (ii) and (iii) for $T$ , we get,

$T = \frac{1}{2} \frac{M m g}{M + 2 m}$

Substituting the value, we get

$T = \frac{1}{2} \frac{(2) (0.5) (9.8)}{2 + (0.5) (2)} = 1.63 N$

(b) From Eq. (iii), angular retardation of drum

$α = \frac{2 T}{M R} = \frac{(2) (1.63)}{(2) (0.2)} = 8.15 r a d / s^{2}$

or linear retardation of block

$a = R α = (0.2) (8.15) = 1.63 m / s^{2}$

At the moment when angular velocity of drum is

$ω_{0} = 10 r a d / s$

The linear velocity of block will be

$v_{0} = ω_{0} R = (10) (0.2) = 2 m / s$

Now, the distance $(s)$ travelled by the block until it comes to rest will be given by

$\begin{aligned} s & = \frac{v_{0}^{2}}{2 a} (Using v^{2} = v_{0}^{2} - 2 a s with v = 0) \\ = \frac{(2)^{2}}{2 (1.63)} or s = 1.22 m \end{aligned}$

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Rotation 5 Question 29

41. A block $X$ of mass $0.5 k g$ is held by a long massless string on a frictionless inclined plane of inclination $30^{\circ}$ to the horizontal. The string is wound on a uniform solid cylindrical drum $Y$ of mass $2 k g$ and of radius $0.2 m$ as shown in figure.

Answer:

इंजीनियरिंग की तैयारी

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Rotation 5 Question 29

41. A block X of mass 0.5kg is held by a long massless string on a frictionless inclined plane of inclination 30∘ to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2kg and of radius 0.2m as shown in figure.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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41. A block $X$ of mass $0.5 k g$ is held by a long massless string on a frictionless inclined plane of inclination $30^{\circ}$ to the horizontal. The string is wound on a uniform solid cylindrical drum $Y$ of mass $2 k g$ and of radius $0.2 m$ as shown in figure.