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Circular and Rotation Motion

A ball of mass {500}{g} tied to one end of string is revolved

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A ball of mass $500 g$ tied to one end of string is revolved in a horizontal circle of radius $10 c m$ with a speed $1 / π r e v / s$ in gravity-free space, then the linear velocity, acceleration and tension in the string will be a. $0.1 m / s, 0.4 m / s^{2}, 0.2 N$ b. $0.1 m / s, 04 m / s^{2}, 0.1 N$ c. $0.2 m / s, 0.4 m / s^{2}, 0.2 N$ d. $0.2 m / s, 0.3 m / s^{2}, 0.2 N$

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A ball of mass

500 g

tied to one end of string is revolved in a horizontal circle of radius

10 c m

with a speed

1 / π r e v / s

in gravity-free space, then the linear velocity, acceleration and tension in the string will be a.

0.1 m / s, 0.4 m / s^{2}, 0.2 N

0.1 m / s, 04 m / s^{2}, 0.1 N

0.2 m / s, 0.4 m / s^{2}, 0.2 N

0.2 m / s, 0.3 m / s^{2}, 0.2 N

6 mins ago

Discuss this question LIVE

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Text solutionVerified

Step 1: Calculate the angular velocity (

ω

)

ω = 2 π N = 2 π \times \frac{1}{π} = 2 r a d / s

Step 2: Calculate the mass (

m

) and radius (

r

)

m = 0.5 k g, r = 0.1 m

Step 3: Calculate the linear velocity (

v

)

v = r ω = 0.1 \times 2 = 0.2 m / s

Step 4: Calculate the linear acceleration (

a

)

a = ω^{2} r = (2)^{2} \times 0.1 = 0.4 m / s^{2}

Step 5: Calculate the tension in the string (

T

)

T = m ω^{2} r = 0.5 \times 0.4 = 0.2 N

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Question Text	A ball of mass $500 g$ tied to one end of string is revolved in a horizontal circle of radius $10 c m$ with a speed $1 / π r e v / s$ in gravity-free space, then the linear velocity, acceleration and tension in the string will be a. $0.1 m / s, 0.4 m / s^{2}, 0.2 N$ b. $0.1 m / s, 04 m / s^{2}, 0.1 N$ c. $0.2 m / s, 0.4 m / s^{2}, 0.2 N$ d. $0.2 m / s, 0.3 m / s^{2}, 0.2 N$
Topic	Circular and Rotation Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1

Text solutionVerified

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