36. A stone of mass 1kg is tied to one end of a
string of length 0.5 m. It is whirled in a vertical
circle. If the maximum tension in the string is
58.8N, the velocity at the top is
1) 1.82 ms-1
2) 2.2ms-1
3) 3.26 ms-1 4) 2.87 ms-1
Answers
The velocity at the top is 2.2 m/s.
Explanation:
=> The maximum tension in the string at the bottom of the circle,
T = mv² / r + mg
58.8 = 1*v²/0.5 + (1 * 9.8)
v²/0.5 = 58.8 - 9.8
v² = 49 * 0.5
v² = 24.5
=> According to the law of conservation of energy,
The sum of Potential Energy and Kinetic Energy at the top is equals to Kinetic Energy at the bottom.
mg(2r) + 1/2m(v')² = 1/2mv²
g(2r) + 1/2(v')² = 1/2v²
g(4r) + (v')² = v²
(v')² = v² - 4gr
v² = 24.5 - (4 * 9.8 * 0.5)
= 24.5 - 19.6
= 4.9
v = √4.9
v = 2.21 m/s ≈ 2.2 m/s
Hence, the velocity at the top is 2.2 m/s.
Learn more:
Q:1 A stone of mass 1kg is tied to the end of a string of 1 m length. It is whirled in a vertical circle . If the velocity of the stone at the top be 4m/s. What is the tension in the string (at the instant ) ?
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Q:2 Object of mass 1 kg is tied to one end of a string of length 9 metre and whirled in a vertical circle what is the minimum speed required at the lowest position in a circle.
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Answer:The maximum tension in the string at the bottom of the circle,
T = mv² / r + mg
58.8 = 1*v²/0.5 + (1 * 9.8)
v²/0.5 = 58.8 - 9.8
v² = 49 * 0.5
v² = 24.5
= According to the law of conservation of energy,
The sum of Potential Energy and Kinetic Energy at the top is equals to Kinetic Energy at the bottom.
mg(2r) + 1/2m(v')² = 1/2mv²
g(2r) + 1/2(v')² = 1/2v²
g(4r) + (v')² = v²
(v')² = v² - 4gr
v² = 24.5 - (4 * 9.8 * 0.5)
= 24.5 - 19.6
= 4.9
v = √4.9
v = 2.21 m/s ≈ 2.2 m/s Ans...
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Explanation: