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A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm.
Correct answer is '5'. Can you explain this answer?
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A ring rolls on a horizontal surface without sliding. The velocity of ...
Conserving angular momentum about point O, we get
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A ring rolls on a horizontal surface without sliding. The velocity of ...
Problem:
A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'.
Solution:
The solution involves the conservation of energy and momentum. The ring rolls without slipping, which means that its linear velocity is equal to the product of its angular velocity and radius. Therefore, we have
v = ωR
Where,
v = velocity of center of the ring
ω = angular velocity of the ring
R = radius of the ring
Step 1: Calculate the velocity of the ring just before the impact
v = ωR
100 = ω(17)
ω = 100/17
ω = 5.88 rad/s
Therefore, the velocity of the ring just before the impact is 5.88 rad/s.
Step 2: Calculate the velocity of the ring just after the impact
The ring encounters a step of height 0.3R. Therefore, the potential energy of the ring just before the impact is mgh = m(0.3R)g, where m is the mass of the ring and g is the acceleration due to gravity.
The ring will lose potential energy as it goes up the step, and gain kinetic energy as it comes down the step. Therefore, we can use the law of conservation of energy to find the velocity of the ring just after the impact.
mgh = (1/2)mv^2 + (1/2)Iω^2
Where,
m = mass of the ring
h = height of the step
g = acceleration due to gravity
v = velocity of the center of the ring just after the impact
I = moment of inertia of the ring about its center of mass
ω = angular velocity of the ring just after the impact
The moment of inertia of a ring about its center of mass is given by I = (1/2)mr^2, where r is the radius of the ring.
Substituting the values, we get
m(0.3R)g = (1/2)mv^2 + (1/2)(1/2)mr^2ω^2
Simplifying, we get
v^2 = (3/2)g(0.3R) + (1/2)r^2ω^2
Substituting the values, we get
v^2 = (3/2)(9.81)(0.3)(17) + (1/2)(17)^2(5.88)^2
v^2 = 457.23
v = 21.37 cm/s
Therefore, the velocity of the ring just after the impact is 21.37 cm/s.
Step 3: Calculate the angular velocity of the ring just after the impact
v = ωR
21.37 = ω(17)
ω = 21.37/17
ω = 1.26 rad/s
Therefore, the angular
A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'.
Solution:
The solution involves the conservation of energy and momentum. The ring rolls without slipping, which means that its linear velocity is equal to the product of its angular velocity and radius. Therefore, we have
v = ωR
Where,
v = velocity of center of the ring
ω = angular velocity of the ring
R = radius of the ring
Step 1: Calculate the velocity of the ring just before the impact
v = ωR
100 = ω(17)
ω = 100/17
ω = 5.88 rad/s
Therefore, the velocity of the ring just before the impact is 5.88 rad/s.
Step 2: Calculate the velocity of the ring just after the impact
The ring encounters a step of height 0.3R. Therefore, the potential energy of the ring just before the impact is mgh = m(0.3R)g, where m is the mass of the ring and g is the acceleration due to gravity.
The ring will lose potential energy as it goes up the step, and gain kinetic energy as it comes down the step. Therefore, we can use the law of conservation of energy to find the velocity of the ring just after the impact.
mgh = (1/2)mv^2 + (1/2)Iω^2
Where,
m = mass of the ring
h = height of the step
g = acceleration due to gravity
v = velocity of the center of the ring just after the impact
I = moment of inertia of the ring about its center of mass
ω = angular velocity of the ring just after the impact
The moment of inertia of a ring about its center of mass is given by I = (1/2)mr^2, where r is the radius of the ring.
Substituting the values, we get
m(0.3R)g = (1/2)mv^2 + (1/2)(1/2)mr^2ω^2
Simplifying, we get
v^2 = (3/2)g(0.3R) + (1/2)r^2ω^2
Substituting the values, we get
v^2 = (3/2)(9.81)(0.3)(17) + (1/2)(17)^2(5.88)^2
v^2 = 457.23
v = 21.37 cm/s
Therefore, the velocity of the ring just after the impact is 21.37 cm/s.
Step 3: Calculate the angular velocity of the ring just after the impact
v = ωR
21.37 = ω(17)
ω = 21.37/17
ω = 1.26 rad/s
Therefore, the angular
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A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer?
Question Description
A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer?.
A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ring rolls on a horizontal surface without sliding. The velocity of its center is v. It encounters a step of height 0.3R, where R is the radius of the ring. Calculate the angular velocity (in rad/s) of the ring just after the impact. Assume that the ring does not return back and there is sufficient friction to avoid slipping. Take v = 100 cm/s, R = 17 cm. Correct answer is '5'. Can you explain this answer?.
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