A disk of radius 12 cm, initially at rest, begins rotating about its axis with a constant angular acceleration of 7.8 rad/s2. What are the following values at t = 4.0 s?
(a) the angular velocity of the disk
(b) the tangential acceleration at and the centripetal acceleration ac of a point at the edge of the disk
at =
ac =
I don't know how to find ac, but I can help you with at and angular velocity of the disk.
(a) (7.8 rad/s^2)/4.0s = 31.2=angular velocity
at=7.8*.12=0.936
Tangential acceleration is the instantaneous linear acceleration associated with the particle at distance (r) from the center of rotation.
thus
Centripetal Accel= (angular accel)(r)
ac = (omega^2)*(radius)
= (31.2^2)*(0.12 m)
= 116.8
(a) The angular velocity of the disk can be found using the formula:
angular velocity (ω) = initial angular velocity + (angular acceleration * time)
Since the disk is initially at rest, the initial angular velocity is 0 rad/s. Plugging in the given values:
ω = 0 + (7.8 rad/s² * 4.0 s)
ω = 31.2 rad/s
So, the angular velocity of the disk at t = 4.0 s is 31.2 rad/s.
(b) To find the tangential acceleration at the edge of the disk, we can use the formula:
tangential acceleration (at) = radius * angular acceleration
Given that the radius (r) of the disk is 12 cm = 0.12 m, and the angular acceleration (α) is 7.8 rad/s²:
at = 0.12 m * 7.8 rad/s²
at = 0.936 m/s²
So, the tangential acceleration at the edge of the disk at t = 4.0 s is 0.936 m/s².
The centripetal acceleration (ac) at the edge of the disk is equal to the tangential acceleration since they both point towards the center of the circular motion. Therefore, ac = 0.936 m/s².
To find the values at t = 4.0 s, we need to use the equations of rotational motion.
(a) The angular velocity of the disk can be found using the equation:
ω = ω0 + αt
where ω is the final angular velocity, ω0 is the initial angular velocity (which is zero in this case since the disk is initially at rest), α is the constant angular acceleration, and t is the time.
Plugging in the given values, we have:
ω = 0 + (7.8 rad/s^2)(4.0 s)
= 31.2 rad/s
So, the angular velocity of the disk at t = 4.0 s is 31.2 rad/s.
(b) The tangential acceleration (at) of a point at the edge of the disk can be found using the equation:
at = αr
where r is the radius of the disk.
Plugging in the given values, we have:
at = (7.8 rad/s^2)(12 cm)
= 93.6 cm/s^2
So, the tangential acceleration at the edge of the disk at t = 4.0 s is 93.6 cm/s^2.
The centripetal acceleration (ac) of a point at the edge of the disk is given by:
ac = ω^2r
Plugging in the values we have:
ac = (31.2 rad/s)^2(12 cm)
= 11982.24 cm/s^2
So, the centripetal acceleration at the edge of the disk at t = 4.0 s is 11982.24 cm/s^2.