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A turntable rotates counterclockwise at 78 rpm. A speck of dust on the turntable is at an initial angular position of 0.45 radians at t = 0 seconds. What is the angular position of the speck at t = 8.8 seconds relative to a reference angular position of 0 radians (your answer should be between 0 and 2\piπ radians, round answer to one decimal)?

User Valerion
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1 Answer

3 votes

Answer:

Displacement will be 11 revolution and 1.02π radian

Explanation:

We have given that initial angular position = 0.45 radian

Angular speed
\omega =78rpm=(2* \pi*  78)/(60)=8.164rad/sec

Time is given as t = 8.8 sec

So angular displacement in 8.8 sec


\Theta =8.8* * 8.164=71.8432radian

As he has already covered 0.45 radian

So total angular displacement = 0.45 + 71.8432 = 72.2932 radian

We know that 1 revolution = 2π = 2×3.14 = 6.28 radian

So 11 revolution = 11×6.28 = 69.08 radian

Left dispalcement = 72.2932 - 69.08 = 3.2132 radian = 1.02π radian

User Manthrax
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