I\'m having a little problem solving this problem: Can you please break it down

Question

I'm having a little problem solving this problem: Can you please break it down step by step please...
A solid disc of 4.00 m diameter is rotating around its axes through its center, has an initial angular velocity of 2.60 rev/min. A constant angular acceleration equal to 1.800 rad/s2 is acting on the wheel, during the time, t = 4.20 s.
(a) What is its angular velocity at the end of 4.20 sec?
(b) Through what angle in degrees has the wheel turned between t = 0 and t = 4.20 s?
(c)What is its linear velocity of the point on the rim at the end of 4.20 sec?

Explanation / Answer

The given initial angular velocity is the Greek letter i = 2.60 rev/min.

The angular acceleration is the Greek letter = 1.800 rad/s2 .

PART A>>>>      We are asked to find the final angular velocity, f , after 4.20 sec .

The rotational kinematics eqn is    f = i + t     we cannot yet plug into this eqn since and do not have consistent units.   Let's convert the 2.6 rev/min into rad/sec

(2.6 rev /min ) ( 2pi rad /1 rev) ( 1 min / 60 sec ) = 0.0867pi  rad/sec      

now plug into the above kinematics eqn:  f       = 0.0867 pi rad/sec + (1.8 rad/s2) (4.20sec)   = 7.83 rad/sec   final answer

PART B>>>>   the kinematics eqn for finding the angle theta is:

= i t + 1/2    t2     plug into this

= (0.0867 pi rad/sec)(4.2 sec) + 1/2 (1.8 rad/s2 ) (4.2 s)2 =   17.02 radians   Now we must convert this result to degrees since the answer was requested to be in degrees.

(17.02 rad) (180 deg / pi rad) = 975.2 degrees final answer

PART C>>>>   linear velocity v is related to angular velocity by the eqn   v = r , where r is the radius.

r= 2.00 m in this case.       so,   v = (2.0 m) (7.83 rad/sec) =   15.66 m/s   final answer

                           

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