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# Centripetal acceleration and tangential acceleration?

A door in a hospital has a pneumatic closer that pulls the door shut such that the doorknob moves with contant speed over most of its path. In this part of its motion:

a) Does the doorknob experience a centripetal acceleration?

b) Does it experience a tangential acceleration?

Hurrying to an emergency, a nurse gives the door a sharp push to the closed door. The door swings open against the pneumatic device, slowing down and then reversing its motion. At the moment the door is open the widest:

c) does the doorknob have a centripetal acceleration?

d)Does it have a tangential acceleration?

Relevance

Well, for this door situation, it wasn't necessarily stated that it is a rotating door (it could be a sliding door instead). But I will assume it is a rotating door.

A) Yes, the doorknob does experience centripetal acceleration. It is traveling at a constant speed, but it is changing direction. Changing direction while still moving, requires a centripetal acceleration, and is directed toward the door hinge (hence the name).

B) No, The doorknob is a fixed distance from the stationary hinge. The doorknob moves at a constant speed. No change in speed means no acceleration tangent to the path of motion.

C) No. There is no centripetal acceleration, because the doorknob is instantaneously stationary when open at its maximum. Zero velocity = zero centripetal acceleration (formula is a=v^2/r, and v=0) required to change the direction. All acceleration at this point is in the direction of the doorknob's travel direction, and not perpendicular toward the hinge.

D) Yes. The doorknob is reversing at this moment. It is slowing down, then speeding up. The doorknob is undergoing acceleration along its direction of motion, which is the meaning of tangential acceleration.

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• a)

Since the knob moves with constant speed along an arc of a circle, the acceleration is centripetal acceleration.

b)

Since it has a constant speed it has no tangential acceleration.

C)

Since its instantaneous velocity is zero, it has no centripetal acceleration.

d)

Since it begins its reverse motion from rest, it has a tangential acceleration at that instant.

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• Let w = angular speed, rad/sec Then, at some time t = T, w^2*r = r*(dw/dt), r = radius of bit So, w^2 = 2dw/dt Let a = angular accel. = constant, rad/s^2 w = aT Then, (aT)^2 = 2*(dw/dt) = 2a a = 2/T^2 Let total angle traveled = theta = (a*T^2)/2 Then theta = (a*T^2)/2 = (T^2/2)*(2/T^2) = 1 radian Answer

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