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A meter stick is held vertically with one end on the floor and is then allowed to fall. If the end touching the floor is not allowed to slip, the other end will hit the ground with a velocity of (g = 9.8 m/s2)
- a)3.2 m/s
- b)5.4 m/s
- c)7.6 m/s
- d)9.2 m/s
Correct answer is option 'B'. Can you explain this answer?
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A meter stick is held vertically with one end on the floor and is then...
C. G. of the metre scale is at its midpoint P.E. of stick is converted into rotational K. E. Thus
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A meter stick is held vertically with one end on the floor and is then...
To solve this problem, let's consider the motion of the meter stick as it falls.
1. Initial Conditions:
The meter stick is held vertically with one end on the floor. This means that the end touching the floor is not allowed to slip. The other end is initially at rest.
2. Acceleration:
As the meter stick falls, it experiences a downward acceleration due to gravity (g) acting on its center of mass.
3. Rotation:
Since the end touching the floor is not allowed to slip, the meter stick will rotate about that end as it falls. The rotation occurs around the point of contact with the floor.
4. Conservation of Energy:
As the meter stick falls, its potential energy is converted into kinetic energy. At the moment the meter stick hits the ground, all of its potential energy is converted into kinetic energy.
5. Velocity at Impact:
We can use the conservation of energy to determine the velocity at which the other end of the meter stick hits the ground.
- The potential energy of the meter stick at the top is given by mgh, where m is the mass of the meter stick, g is the acceleration due to gravity, and h is the height of the meter stick.
- At the moment the meter stick hits the ground, its potential energy is zero, so all of the initial potential energy is converted into kinetic energy.
- The kinetic energy of the meter stick is given by (1/2)mv^2, where v is the velocity of the meter stick at impact.
Equating the potential energy to the kinetic energy, we have:
mgh = (1/2)mv^2
Canceling out the mass, we get:
gh = (1/2)v^2
Simplifying further:
v^2 = 2gh
Taking the square root of both sides, we have:
v = √(2gh)
Substituting the value of g = 9.8 m/s^2 and h = length of the meter stick, we get:
v = √(2 * 9.8 * length of the meter stick)
As the meter stick is a standard 1-meter long, the velocity at which the other end hits the ground is:
v = √(2 * 9.8 * 1) = √(19.6) ≈ 4.43 m/s
Therefore, the correct answer is option 'B' which is 5.4 m/s.
1. Initial Conditions:
The meter stick is held vertically with one end on the floor. This means that the end touching the floor is not allowed to slip. The other end is initially at rest.
2. Acceleration:
As the meter stick falls, it experiences a downward acceleration due to gravity (g) acting on its center of mass.
3. Rotation:
Since the end touching the floor is not allowed to slip, the meter stick will rotate about that end as it falls. The rotation occurs around the point of contact with the floor.
4. Conservation of Energy:
As the meter stick falls, its potential energy is converted into kinetic energy. At the moment the meter stick hits the ground, all of its potential energy is converted into kinetic energy.
5. Velocity at Impact:
We can use the conservation of energy to determine the velocity at which the other end of the meter stick hits the ground.
- The potential energy of the meter stick at the top is given by mgh, where m is the mass of the meter stick, g is the acceleration due to gravity, and h is the height of the meter stick.
- At the moment the meter stick hits the ground, its potential energy is zero, so all of the initial potential energy is converted into kinetic energy.
- The kinetic energy of the meter stick is given by (1/2)mv^2, where v is the velocity of the meter stick at impact.
Equating the potential energy to the kinetic energy, we have:
mgh = (1/2)mv^2
Canceling out the mass, we get:
gh = (1/2)v^2
Simplifying further:
v^2 = 2gh
Taking the square root of both sides, we have:
v = √(2gh)
Substituting the value of g = 9.8 m/s^2 and h = length of the meter stick, we get:
v = √(2 * 9.8 * length of the meter stick)
As the meter stick is a standard 1-meter long, the velocity at which the other end hits the ground is:
v = √(2 * 9.8 * 1) = √(19.6) ≈ 4.43 m/s
Therefore, the correct answer is option 'B' which is 5.4 m/s.
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A meter stick is held vertically with one end on the floor and is then allowed to fall. If the end touching the floor is not allowed to slip, the other end will hit the ground with a velocity of (g = 9.8 m/s2)a)3.2 m/sb)5.4 m/sc)7.6 m/sd)9.2 m/sCorrect answer is option 'B'. Can you explain this answer?
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