A belt is wrapped around the edge of a pulley that is 40 cm in diameter. The
pulley rotates with a constant angular acceleration of 3.50 rad/s². At t = 0, the
rotational speed is 2rad/s. What is the angular displacement and angular
velocity of the pulley 2 s later?
Answers
Answer:
A belt is wrapped around the edge of a pulley that is 40 cm in diameter. The pulley rotates with a constant angular acceleration of 3.50 rad/s(squared).At t=0, the rotational speed is 2 rad/s. What are the angular displacement and angular velocity of the pulley 2 s later.
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Answer:
Explanation:
wf= wi + at
wf=2rad/s+ (3.50rad/s2 x2s)
wf=2rad/s + 7rad/s
wf=9rad/s
ANGULAR DISPLACEMENT
wf^2 - wi^2 = 2a(0) note: (here zero is thetha didnt find a thetha symbol)
(9rad/s)^2 - (2rad/s)^2 = 2a0
81rad2/s2-4rad2/s2=2*3.50rad/s2(0) 0 = angular displacement
(77rad2/s2) / (7rad/s2 ) =0
11rad=0
hence angular displacement and angular velocity after 2s is 11rad and 9 rad/s respectively