A cyclist accelerates from rest at a rate of 1.00 .
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Question “A cyclist accelerates from rest at a rate of 1.00 .”
A cyclistaccelerates from rest at a rate of 1.00 .
How fast will a point on the rim of the tire ( = 68 ) at the top be moving after 3.7 ? [Hint: At any moment, the lowest point on the tire is in contact with theground and is at rest – see the figure .]
Answer
Answer 1
7
Answer 2
Velocity at the tire’s center after 2.6 seconds = 0 + 2/6 = 2.6 m/sec
The distance between the top and the center of the tire is twice that of its diameter.
As v = wr
The angular velocity for the tire w is equal.
The distance from the top point to the center of the tire is twice as far as the distance from ther center.
Velocity at the top will be twice that velocity = 2*2.6 = 5 m/sec
Answer 3
A cyclist accelerates from rest at a rate of 1.00 m/s2
. How fast will a point at the top of the rim of the tire (diameter = 68 cm
) be moving after 3.0 s?
Conclusion
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