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Centripetal Acceleration and Gravity - by Mr. Paul Crosa, 2001

Centripetal Acceleration

1. A girl sits on a tire that is attached to an overhanging tree limb by
a rope.  The girl's father pushes her so that her centripetal
acceleration is 3.0 m/s^2.  If the length of the rope is 2.1 m, what is the girl's
tangential speed?

Here's what you know, centripetal A = 3.0 m/s^2 , and r = 2.1 m.
Use the formula ac = vt2/r

1. v is your tangential velocity, so find that

2. find v by making the equation read (ac *r)^(1/2)=vt

3. plug-in the values (3 m/s^2*2.1 m)^(1/2) = 2.51 m/s

2. A young boy swings a yo-yo horizontally above his head so that the
yo-yo has a centripetal acceleration of 250 m/s^2.  If the yo-yo's
string is .5 m long, what is the yo-yo's tangential speed?

Here's what you know, centripetal A = 250 m/s^2 , and r = .5
m.  Use the formula ac = vt2/r

1. v is your tangential velocity, so find that

2. find v by making the equation read (ac*r)^(1/2)=vt

3. plug-in the values (250 m/s^2*.5 m)^(1/2) = 11.18 m/s

3. A dog sits 1.5 m from the center of a merry-go-round. If the dog
undergoes a 1.5 m/s^2 centripetal acceleration, what is the dog's linear
speed? What is the angular speed of the merry go round?

Here's what you know, centripetal A = 1.5 m/s^2 , and r = 1.5
m.  Use the formula ac = vt2/r and w = vt / r

1. v is your tangential or linear velocity, so find that first

2. find v by making the equation read (ac*r)^(1/2)=vt

3. plug-in the values (1.5 m/s^2*1.5 m)^(1/2) = 1.5 m/s

4. Next use the equation w = vt / r

5. Plug-in the values and you get 1.5 m/s / 1.5 m = 1.0 rad/s

4. A race car moves along a circular track at an angular speed of .512
rad/s.  If the car's centripetal acceleration is 15.4 m/s^2, what is the
distance between the car and the center of the track?

Here's what you know, centripetal A = 15.4 m/s^2 , and w = .512
rad/s.  Use the formula ac = vt2/r and w = vt / r

1. first set ac = w*v and find v

2. 15.4 m/s^2 / .512 rad/s = 30 m/s

3. use v to find r in the equation w = vt / r rewriting it as r=vt / w

4. plug-in the values and you get 30 m/s / .512 rad/s = 58.6 m

5. A piece of clay sits 0.20 m from the center of a potter's wheel.  If
the potter spins the wheel at an angular speed of 20.5 rad/s, what is
the magnitude of the centripetal acceleration of the piece of clay on the
wheel?

Here's what you know, r = .2 m , and w = 20.5 rad/s.  Use the
formula ac = vt2/r and w = vt / r

1. use the equation w = vt / r to find v rewriting it as w*r = v

2. plug-in the values and you get 20.5 rad/s * .2 m = 4.1 m/s

3. plug the velocity you just got into ac = vt2/r

4. plug-in the values to get (4.1 m/s)^2 / .2 m = 84.05 m/s^2