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24. I FIGURE EX4.24 shows the angular-position-versus-time graph for a particle moving in a circle. What is the particle's angular velocity at (a) t=1s, (b) t=4s, and (c) t=7s?

Short Answer

Expert verified

Part (a). The angular velocity of the particle at t=1sis role="math" localid="1652184311109" 6.28rad/s.

Part (b). The angular velocity of the particle att=4s is 0โ€‰โ€Šrad/s.

Part (c). The angular velocity of the particle att=7s isโˆ’6.28rad/s .

Step by step solution

01

Part (a)

The angular velocity of the particle is,

ฯ‰=ฮธ1โˆ’ฮธ0t1โˆ’t0

At t1=1sthe angular position of the particle isฮธ1=2ฯ€

At t0=0sthe initial angular position of the particle isฮธ0=0

The angular velocity of the particle is,

ฯ‰=(2ฯ€rad)โˆ’0rad1sโˆ’0s=6.28rad/s

The angular velocity of the particle att=1s is6.28rad/s

02

Part (b)

The angular velocity of the particle is,

ฯ‰=ฮธ4โˆ’ฮธ2t4โˆ’t2

The particle at t4=4s, the angular position is ฮธ4=4ฯ€

The particle at t2=2s, the initial angular position isฮธ2=4ฯ€

The angular velocity of the particle is,

ฯ‰=(4ฯ€rad)โˆ’(4ฯ€rad)4sโˆ’2s=0rad/s

Thus, the angular velocity of the particle att=4s is0rad/s .

03

Part (c)

The angular velocity of the particle is,

ฯ‰=ฮธ7โˆ’ฮธ6t7โˆ’t6

At t7=7stime the angular position of the particle isฮธ7

At t6=6s.time the initial angular position of the particle isฮธ6

The angular velocity of the particle is,

ฯ‰=(0rad)โˆ’(2ฯ€rad)7sโˆ’6s=โˆ’2ฯ€rad/s=โˆ’6.28rad/s

Thus, the angular velocity of the particle at t=7sisโˆ’6.28rad/s

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