A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
Practice Questions
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Q1
A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
(2/5)m(10^2)
(1/2)m(10^2)
(1/3)m(10^2)
(5/2)m(10^2)
Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).
Questions & Step-by-step Solutions
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Q
Q: A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
Solution: Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).
Steps: 10
Step 1: Understand the problem. A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom from a height of 5 m.
Step 2: Identify the energy types involved. The ball has potential energy (PE) at the top and kinetic energy (KE) at the bottom.
Step 3: Write the formula for potential energy: PE = mgh, where m is mass, g is acceleration due to gravity (approximately 9.81 m/s²), and h is height (5 m).
Step 4: Write the formula for kinetic energy: KE = (1/2)mv² + (1/2)I(ω²), where v is the linear speed (10 m/s) and I is the moment of inertia.
Step 5: Since the ball is rolling, relate linear speed (v) to angular speed (ω) using the formula: ω = v/r, where r is the radius of the sphere.
Step 6: For a solid sphere, the moment of inertia I is given by the formula: I = (2/5)mr².
Step 7: Use conservation of energy: Potential energy at the top = Kinetic energy at the bottom. So, mgh = (1/2)mv² + (1/2)I(ω²).
Step 8: Substitute the known values into the equation: mgh = (1/2)m(10²) + (1/2)(2/5)m(10²).
Step 9: Simplify the equation to find I. You will find that I = (2/5)m(10²).
Step 10: Conclude that the moment of inertia I can be expressed in terms of mass m and the speed of the ball.
