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The bar of Prob. 5/82 is repeated here. The ends of the 0.4-m bar remain in contact with their re- spective support surfaces. End B has a velocity of 0.5 m/s and an acceleration of 0.3 m/s2 in the di- rections shown. Determine the angular accelera- tion of the bar and the acceleration of end A.

Respuesta :

curiousmuktar

Answer:

0.24

Explanation:

w=vt

then

w1/v1=w2/v2

0.4/0.5=w2/0.3

w2= 0.3*0.4/0.5=0.24//

abbusaicharan02

the angular acceleration of the bar and the acceleration of end A is 0.24.

w=vt

then

w1/v1=w2/v2

0.4/0.5=w2/0.3

w2= 0.3*0.4/0.5=0.24.

What is angular acceleration?

Angular acceleration is defined as the rate of change of angular velocity over time. Usually expressed in radians / second / second. Therefore, α = dωdt. Angular acceleration is also known as rotational acceleration.

Angular velocity is the rate of change in angular displacement, and angular acceleration is the rate of change in angular velocity.

. Angular acceleration is the change in angular velocity divided by time, and tangential acceleration is the change in linear velocity divided by time.

Angular acceleration is taken directly from the definition of α = ΔωΔt α = ΔωΔt because it gives the final angular velocity and time. You can see that Δω is 250 rpm and Δt is 5.00s. α=ΔωΔt=250U /min5.00 seconds。 α=ΔωΔt=250U /min5.00 seconds。

Learn more about angular acceleration here: https://brainly.com/question/21278452

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