A coin with a radius of 5 cm rolls down a 30° inclined plane in a straight line without slipping. When the coin is at a height of 50 cm above the ground, it's angular speed is 45.0 rad/s. If the moment of inertia of the coin is (1/2) MR2, what will be the velocity of the coin when it reaches the ground ?
A coin with a radius of 5 cm rolls down a 30° inclined plane in a straight line without slipping. When the coin is at a height of 50 cm above the ground, it's angular speed is 45.0 rad/s. If the moment of inertia of the coin is (1/2) MR2, what will be the velocity of the coin when it reaches the ground ?
College Physics
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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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
Transcribed Image Text:A coin with a radius of 5 cm rolls down a 30° inclined plane in a straight line
without slipping. When the coin is at a height of 50 cm above the ground,
it's angular speed is 45.0 rad/s. If the moment of inertia of the coin is
(1/2) MR2, what will be the velocity of the coin when it reaches the ground ?
Expert Solution
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Step 1
Assume that the ground will be the datum point.
On the system, no external force is acting on this. Therefore, from conservation of energy.
The formula for the angular velocity in terms of linear velocity is:
Step 2
Use the expression of the angular velocity, and known values and rewrite the expression of energy as:
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