If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit
1 answer:
Answer:
Explanation:
If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.
Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t
The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t
So the linear speed compared to angular speed is
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The velocity of the particle is given by the derivative of the position vector:
(a) The particle is moving in the <em>x</em>-direction when the <em>y</em>-component of velocity is zero:
But we want <em>t</em> > 0, so this never happens, unless 2<em>c</em> = <em>d</em> is given, in which case the <em>y</em>-component is always zero.
(b) Similarly, the particle moves in the <em>y</em>-direction when the <em>x</em>-component vanishes:
We drop the zero solution, and we're left with
In the case of 2<em>c</em> = d, this times reduces to <em>t</em> = <em>c</em>/(6<em>c</em>) = 1/6.
OPTION D is the correct answer.
Refer to the attachment for complete calculation...
