A uniform rod of length L and mass M is pivoted at one end and released from res

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What’s inside this PDF?

Question: A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?

Options:

√(3g/L)
√(2g/L)
√(g/L)
√(4g/L)

Correct Answer: √(2g/L)

Solution:

Using conservation of energy, potential energy at the top = rotational kinetic energy at the bottom. mgh = (1/2)Iω^2. For a rod, I = (1/3)ML^2, h = L/2. Solving gives ω = √(3g/L).

A uniform rod of length L and mass M is pivoted at one end and released from res

Practice Questions

A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?

√(3g/L)
√(2g/L)
√(g/L)
√(4g/L)

Questions & Step-by-Step Solutions

A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?

Step 1: Understand the problem. We have a uniform rod of length L and mass M that is pivoted at one end and falls due to gravity.
Step 2: Identify the energy types involved. When the rod is at the top, it has potential energy (PE) because of its height. Just before it hits the ground, it has rotational kinetic energy (KE).
Step 3: Write the formula for potential energy. The potential energy at the top is given by PE = mgh, where h is the height of the center of mass of the rod.
Step 4: Find the height (h) of the center of mass. For a uniform rod, the center of mass is at L/2, so h = L/2.
Step 5: Write the formula for rotational kinetic energy. The rotational kinetic energy is given by KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 6: Find the moment of inertia (I) for the rod. The moment of inertia for a rod pivoted at one end is I = (1/3)ML^2.
Step 7: Set the potential energy equal to the rotational kinetic energy. This gives us the equation mgh = (1/2)Iω^2.
Step 8: Substitute the values of h and I into the equation. This gives us: Mg(L/2) = (1/2)(1/3)ML^2ω^2.
Step 9: Simplify the equation. Cancel M from both sides and simplify to get g(L/2) = (1/6)L^2ω^2.
Step 10: Solve for ω. Rearranging gives ω^2 = (3g)/L, so ω = √(3g/L).

Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
Moment of Inertia – A measure of an object's resistance to changes in its rotation, which for a uniform rod pivoted at one end is given by I = (1/3)ML^2.
Angular Velocity – The rate of rotation of an object, which can be calculated using the relationship between potential energy and kinetic energy.

A uniform rod of length L and mass M is pivoted at one end and released from res

What’s inside this PDF?

A uniform rod of length L and mass M is pivoted at one end and released from res

Practice Questions

Questions & Step-by-Step Solutions

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