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A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.
- a)27.5rad
- b)24.6rad
- c)29.3rad
- d)32.4rad
Correct answer is option 'C'. Can you explain this answer?
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A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter ...
Explanation: ω(avg)=[ω()initial+ω(final)]/2 and θ=ωt.
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A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter ...
To calculate the angle through which the disk rotated during the braking period, we can use the concept of rotational kinetic energy.
1. Calculate the initial angular velocity:
The angular velocity (ω) is given in revolutions per minute (rpm). We need to convert it to radians per second (rad/s).
ω = (2π * n) / 60, where n is the number of revolutions per minute.
In this case, n = 350.
ω = (2π * 350) / 60 = 36.67 rad/s
2. Calculate the initial kinetic energy:
The formula for rotational kinetic energy (KE) is:
KE = (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
Given that the square of the radius of gyration (k^2) is 0.2, we can calculate the moment of inertia (I) using the formula:
I = m * k^2, where m is the mass of the disk.
Given that the mass (m) is 1000 kg, the moment of inertia is:
I = 1000 * 0.2^2 = 40 kg·m^2
Substituting the values, we can calculate the initial kinetic energy:
KE = (1/2) * 40 * (36.67)^2 = 26694.33 J
3. Calculate the final kinetic energy:
The final kinetic energy is zero because the disk comes to a stop.
4. Calculate the work done by the brake:
The work done by the brake is equal to the change in kinetic energy. Since the final kinetic energy is zero, the work done by the brake is equal to the initial kinetic energy:
Work = 26694.33 J
5. Calculate the angle through which the disk rotated:
The work done by the brake is given by the equation:
Work = τ * θ, where τ is the torque exerted by the brake and θ is the angle through which the disk rotated.
Since the torque (τ) is constant, we can rearrange the equation to solve for θ:
θ = Work / τ
To find τ, we need to determine the braking time (t) and the moment of inertia (I):
τ = I * α, where α is the angular acceleration.
The angular acceleration can be calculated using the formula:
α = ω / t
Given that the braking time (t) is 1.6 s, we can calculate the angular acceleration:
α = 36.67 / 1.6 = 22.92 rad/s^2
Substituting the values, we can calculate the torque:
τ = 40 * 22.92 = 916.8 N·m
Now we can calculate the angle:
θ = 26694.33 / 916.8 = 29.18 rad
Therefore, the angle through which the disk rotated during the braking period is approximately 29.3 rad (option C).
1. Calculate the initial angular velocity:
The angular velocity (ω) is given in revolutions per minute (rpm). We need to convert it to radians per second (rad/s).
ω = (2π * n) / 60, where n is the number of revolutions per minute.
In this case, n = 350.
ω = (2π * 350) / 60 = 36.67 rad/s
2. Calculate the initial kinetic energy:
The formula for rotational kinetic energy (KE) is:
KE = (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
Given that the square of the radius of gyration (k^2) is 0.2, we can calculate the moment of inertia (I) using the formula:
I = m * k^2, where m is the mass of the disk.
Given that the mass (m) is 1000 kg, the moment of inertia is:
I = 1000 * 0.2^2 = 40 kg·m^2
Substituting the values, we can calculate the initial kinetic energy:
KE = (1/2) * 40 * (36.67)^2 = 26694.33 J
3. Calculate the final kinetic energy:
The final kinetic energy is zero because the disk comes to a stop.
4. Calculate the work done by the brake:
The work done by the brake is equal to the change in kinetic energy. Since the final kinetic energy is zero, the work done by the brake is equal to the initial kinetic energy:
Work = 26694.33 J
5. Calculate the angle through which the disk rotated:
The work done by the brake is given by the equation:
Work = τ * θ, where τ is the torque exerted by the brake and θ is the angle through which the disk rotated.
Since the torque (τ) is constant, we can rearrange the equation to solve for θ:
θ = Work / τ
To find τ, we need to determine the braking time (t) and the moment of inertia (I):
τ = I * α, where α is the angular acceleration.
The angular acceleration can be calculated using the formula:
α = ω / t
Given that the braking time (t) is 1.6 s, we can calculate the angular acceleration:
α = 36.67 / 1.6 = 22.92 rad/s^2
Substituting the values, we can calculate the torque:
τ = 40 * 22.92 = 916.8 N·m
Now we can calculate the angle:
θ = 26694.33 / 916.8 = 29.18 rad
Therefore, the angle through which the disk rotated during the braking period is approximately 29.3 rad (option C).
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Question Description
A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer?.
A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A solid cast iron disk of mass 1000kg is rotating at 350rpm. Diameter of the disk is 1m and time taken to come to stop the disk by brake is 1.6sec.Square of radius of gyration is 0.2. Calculate the angle through which disk rotated during braking period.a)27.5radb)24.6radc)29.3radd)32.4radCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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