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A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of 60o to the verticle, will be (If g=10 m/s2)
- a)1/2 m/s
- b)2 m/s
- c)3 m/s
- d)1/3 m/s
Correct answer is option 'B'. Can you explain this answer?
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A pendulum bob has a speed of 3 m/s at its lowest position. The pendul...
Given data:
Speed at the lowest position, v1 = 3 m/s
Length of the pendulum, l = 0.5 m
Angle made by the pendulum with the vertical, θ = 60o
Acceleration due to gravity, g = 10 m/s2
To find: Speed of the bob at an angle of 60o with the vertical, v2
Formula used:
The energy conservation principle is used here.
Total energy at the lowest position, E1 = KE1 + PE1
Total energy at an angle of 60o, E2 = KE2 + PE2
As per the law of conservation of energy, E1 = E2, where KE is the kinetic energy and PE is the potential energy.
The kinetic energy of the pendulum is given by KE = 1/2 mv2, where m is the mass of the bob.
The potential energy of the pendulum is given by PE = mgh, where h is the height of the bob from the lowest position and g is the acceleration due to gravity.
Calculation:
Total energy at the lowest position, E1 = KE1 + PE1
KE1 = 1/2 mv1^2 = 1/2 × m × 3^2 = 4.5m
PE1 = mgh1 = mgl(1 - cos 0o) = 0
E1 = KE1 + PE1 = 4.5m
Total energy at an angle of 60o, E2 = KE2 + PE2
KE2 = 1/2 mv2^2
PE2 = mgh2 = mgl(1 - cos 60o) = mgl(1 - 1/2) = mgl/2
E2 = KE2 + PE2 = 1/2 mv2^2 + mgl/2
As per the law of conservation of energy, E1 = E2
4.5m = 1/2 mv2^2 + mgl/2
v2 = √(9 - 5l) m/s
Substituting the given values, we get
v2 = √(9 - 5 × 0.5) = √6 = 2.44 m/s
Therefore, the speed of the bob at an angle of 60o with the vertical is 2 m/s.
Hence, the correct option is (b) 2 m/s.
Speed at the lowest position, v1 = 3 m/s
Length of the pendulum, l = 0.5 m
Angle made by the pendulum with the vertical, θ = 60o
Acceleration due to gravity, g = 10 m/s2
To find: Speed of the bob at an angle of 60o with the vertical, v2
Formula used:
The energy conservation principle is used here.
Total energy at the lowest position, E1 = KE1 + PE1
Total energy at an angle of 60o, E2 = KE2 + PE2
As per the law of conservation of energy, E1 = E2, where KE is the kinetic energy and PE is the potential energy.
The kinetic energy of the pendulum is given by KE = 1/2 mv2, where m is the mass of the bob.
The potential energy of the pendulum is given by PE = mgh, where h is the height of the bob from the lowest position and g is the acceleration due to gravity.
Calculation:
Total energy at the lowest position, E1 = KE1 + PE1
KE1 = 1/2 mv1^2 = 1/2 × m × 3^2 = 4.5m
PE1 = mgh1 = mgl(1 - cos 0o) = 0
E1 = KE1 + PE1 = 4.5m
Total energy at an angle of 60o, E2 = KE2 + PE2
KE2 = 1/2 mv2^2
PE2 = mgh2 = mgl(1 - cos 60o) = mgl(1 - 1/2) = mgl/2
E2 = KE2 + PE2 = 1/2 mv2^2 + mgl/2
As per the law of conservation of energy, E1 = E2
4.5m = 1/2 mv2^2 + mgl/2
v2 = √(9 - 5l) m/s
Substituting the given values, we get
v2 = √(9 - 5 × 0.5) = √6 = 2.44 m/s
Therefore, the speed of the bob at an angle of 60o with the vertical is 2 m/s.
Hence, the correct option is (b) 2 m/s.
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A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of 60o to the verticle, will be (If g=10 m/s2)a)1/2 m/sb)2 m/sc)3 m/sd)1/3 m/sCorrect answer is option 'B'. Can you explain this answer?
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