Question

The magnitude of acceleration of a point on the wheels rim at t=1s

Answer 1

The net angular acceleration is 203 .96 m/s^2

Explanation:

The magnitude of acceleration of a point on the wheels rim at t= 1s whose radius is 2m and is spinning with angular velocity 'w' that varies with time as w = 10t^2a. 40root 26m/s^2b. 40 m/s^2c. Zerod. 20m/s^2?

There are two acceleration is circular motion.

Rα = along x- axis

Rω^2 = along y-axis

α T => Rα = R dw/dt = Rd [10t^2]/dt

=> R x 20t = 20Rt = 20 x 2 x 1 = 40 m/s^2

αR = Rω^2 = 2 x (10t^2)2

    = 2 x 100 x t^4 = 200 x (1)^4 = 200 m/s^2

αnet = √ αR^2 + αT^2

       = √  (200)^2 + (40)^2

      = √ 40000 + 1600

      = √ 41600 = 203 .96 m/s^2

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Answer 2

Answer:

Hope this helps u ...