PRACTICE IT A wheel rotates with a constant angular acceleration of 3.25 rad/s 2

Question

PRACTICE IT

A wheel rotates with a constant angular acceleration of 3.25 rad/s2. Assume the angular speed of the wheel is 2.20 rad/s at ti = 0.

(a) Through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and revolutions.
___________ rad
___________ rev

(b) What is the angular speed of the wheel at t = 2.00 s?
___________rad/s

(c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?
___________ rev

EXERCISE

(Use the values from above)

(a) Find the angle through which the wheel rotates between t = 2.00 s and t = 3.80 s.
_______________ rad

(b) Find the angular speed when t = 3.80 s.
____________ rad/s

(c) What is the magnitude of the angular speed three revolutions following t = 3.80 s?
____________ rad/s

Explanation / Answer

(A) theta = wi t + alpha t^2 / 2

theta = (2.20 x 2) + (3.25 x 2^2 / 2)

theta = 10.9 rad .....Ans

revolutions = theta / 2pi = 1.73 rev .....Ans

(B) wf = wi + alpha t

w = 2.20 + (3.25 x 2)

w = 8.7 rad/s .....Ans

(C) that means angular acceleration will be doubled.

alpha = 2 x 3.25= 6.5 rad/s^2

theta = (2 x 2.20 x 2 ) + (6.5 x 2^2 / 2)

= 21.8 rad  

or 3.47 rev ....Abs

EXERCISE:

(A) t = 3.80 - 2 = 1.80 s

theta = (8.7 x 1.80) + (3.25 x 1.80^2 / 2)

= 20.9 rad

(B) w = 8.7 + (3.24 x 1.80) = 14.6 rad/s

(C) w^2 - 14.6^2 = 2(3 x 2pi) (3.25)

w= 18.3 rad/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.