A thin circular hoop is released from rest at the top of a plane inclined at 23.5 degrees above the horizontal. Assume perfect rolling takes place.How long does it take the hoop to roll 8.8 m down the plane surface?Express your answer in s, to at least one digit after the decimal point.
Question
A thin circular hoop is released from rest at the top of a plane inclined at 23.5 degrees above the horizontal. Assume perfect rolling takes place.How long does it take the hoop to roll 8.8 m down the plane surface?Express your answer in s, to at least one digit after the decimal point.
Solution
To find the time it takes for the hoop to roll 8.8 m down the inclined plane, we can use the equations of motion for rolling objects.
First, we need to determine the acceleration of the hoop down the plane. The component of gravity acting parallel to the plane can be found by multiplying the acceleration due to gravity (9.8 m/s^2) by the sine of the angle of inclination (23.5 degrees).
Acceleration = 9.8 m/s^2 * sin(23.5 degrees)
Next, we can use the equation of motion for rolling objects:
Distance = (1/2) * acceleration * time^2
Rearranging the equation to solve for time, we have:
Time = sqrt((2 * Distance) / acceleration)
Plugging in the given values, we have:
Time = sqrt((2 * 8.8 m) / (9.8 m/s^2 * sin(23.5 degrees)))
Calculating this expression will give us the time it takes for the hoop to roll 8.8 m down the inclined plane.
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