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The shaft of a pump starts from rest and has an angular acceleration of 4.62 rad/s2 for 15.4 s. at the end of this interval, what is (a) the shaft's angular speed and (b) the angle through which the shaft has turned?

User Dened
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Final answer:

The angular speed of the shaft is 71.228 rad/s and the angle through which it has turned is 1126.007 rad.

Step-by-step explanation:

The angular speed and the angle through which the shaft has turned can be calculated using the equations of rotational motion. The angular speed can be calculated using the formula:

ω = ω0 + αt

where ω is the final angular speed, ω0 is the initial angular speed (0 rad/s as the shaft starts from rest), α is the angular acceleration (4.62 rad/s²), and t is the time interval (15.4 s). Plugging in the values, we get:

ω = 0 + (4.62)(15.4) = 71.228 rad/s (a)

The angle through which the shaft has turned can be calculated using the formula:

θ = ω0t + (1/2)αt²

where θ is the angle, ω0 is the initial angular speed, α is the angular acceleration, and t is the time interval. Plugging in the values, we get:

θ = 0 + (1/2)(4.62)(15.4)² = 1126.007 rad (b)

User Chanchal Raj
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