A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through

Question

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the constant rate (alpha)1 through the first 400 rad and then losing angular speed at the constant rate - (alpha)1 until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s2.
.
(a) What is the least time required for the rotation?

(b) What is the corresponding value of ?1?

The angular acceleration of a wheel is (alpha) = 6.0t^4? 4.0t^2 with (alpha) in radians per second-squared and t in seconds. At time t = 0, the wheel has an angular velocity of +2 rad/s and an angular position of +1 rad. Write expressions for:

(a) The angular velocity (rad/s)

(b) The angular position (rad) as functions of time (s).

Explanation / Answer

1.
By ?(final)^2 = ?(initial)^2 + 2??
=> ?(final)^2 = 0 + 2 x ? x 400
=>?(final)^2 = 800? ---------------(i)

given a(max) = v^2/r = 400

=>v(max) = ?400r = ?[400 x 0.25] = 10 m/s
=>?(max) = v(max)/r = 10/0.25 = 40 rad/sec ------------(ii)
=>By (i) & (ii) :-
=>(40)^2 = 800?
=>? = 2 rad/sec^2

=>By :- ?(final) = ?(initial) + ?t1
=>40 = 0 + 2 x t1
=>t1 = 20 sec
=>T = 2t1= 2 x 20 = 40 sec

2.
(a)
? = 6.0t4 - 4.0t2
angular velocity, ? = 6/5 t5 - 4/3 t3 + C
at t = 0, ? = 2
2 = C
angular velocity, ? = 6/5 t5 - 4/3 t3 + 2

(b)
? = 6/5 t5 - 4/3 t3 + 2
angular position, x(t) = 1/5 t6 - 1/3 t4 + 2t + C
at t = 0, x(t) = 1
1 = C
angular position, x(t) = 1/5 t6 - 1/3 t4 + 2t + 1

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