# 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 =

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525249

Created
April 4, 2011 3:56pm UTC

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1. 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

525778

Created
April 5, 2011 1:59pm UTC

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2. 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)

526583

Created
April 6, 2011 5:46pm UTC

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= (31.2^2)*(0.12 m)
= 116.8

526838

Created
April 6, 2011 11:07pm UTC

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4. (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)

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.