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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)?

Respuesta :

Answer:

Displacement will be 11 revolution and 1.02π radian

Step-by-step explanation:

We have given that initial angular position = 0.45 radian

Angular speed [tex]\omega =78rpm=\frac{2\times \pi\times  78}{60}=8.164rad/sec[/tex]

Time is given as t = 8.8 sec

So angular displacement in 8.8 sec

[tex]\Theta =8.8\times \times 8.164=71.8432radian[/tex]

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

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