• sushmarao
In summary, the problem involves finding the minimum time required for a suitcase to go around once on a carousel at an airport, given the radius of the circle, coefficient of static friction, and angle. The equations used are v= 2*pi*r/T and acceleration centripetal = v^2/r. However, the free body diagram and resulting equation for problem 1 are incorrect. In problem 2, the linear velocity v is not the same for both places on the rotating disk, so the equation cannot be solved using only that value. Instead, another quantity that is the same for both places must be used.
sushmarao

Homework Statement

#1
The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 14.0 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.860, and the angle θ in the drawing is 26.7 °. What is the minimum time required for your suitcase to go around once?

here is the picture that they included with the problem
http://edugen.wiley.com/edugen/cours...5/ch05p_26.gif

#2
A computer is reading data from a rotating CD-ROM. At a point that is 0.0144 m from the center of the disk, the centripetal acceleration is 155 m/s2. What is the centripetal acceleration at a point that is 0.0845 m from the center of the disc?

Homework Equations

for #1 the equations I've been trying are
v= 2*pi*r/T
and acceleration centripetal = v^2/r

the same equation was used in #2.

the only other equations they taught us in lecture is F=mv^2/r

The Attempt at a Solution

for problem 1 I did a free body diagram and came up with this equation:
(.86)(mg)(cos26.7) = mv^2/14
then the mass cancels
leaving (.86)(9.8)(cos26.7)=v^2/14
so then i solved for v
and then plugged v into
v=2*pi*14/T
and solved for T
but this is not the right answer and I'm totally confused as to what I'm doing wrong!

for #2
the only thing I can think of doing is solving for v^2 by doing
155 = v^2/.0144
and then plugging v^2 into
acceleration centripetal = v^2/.0845

Last edited by a moderator:
Hi sushmarao,

sushmarao said:

Homework Statement

#1
The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 14.0 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.860, and the angle θ in the drawing is 26.7 °. What is the minimum time required for your suitcase to go around once?

here is the picture that they included with the problem
http://edugen.wiley.com/edugen/cours...5/ch05p_26.gif

#2
A computer is reading data from a rotating CD-ROM. At a point that is 0.0144 m from the center of the disk, the centripetal acceleration is 155 m/s2. What is the centripetal acceleration at a point that is 0.0845 m from the center of the disc?

Homework Equations

for #1 the equations I've been trying are
v= 2*pi*r/T
and acceleration centripetal = v^2/r

the same equation was used in #2.

the only other equations they taught us in lecture is F=mv^2/r

The Attempt at a Solution

for problem 1 I did a free body diagram and came up with this equation:
(.86)(mg)(cos26.7) = mv^2/14

I cannot see the picture in the link you gave, but I don't believe this is correct. What does your force diagram look like?

then the mass cancels
leaving (.86)(9.8)(cos26.7)=v^2/14
so then i solved for v
and then plugged v into
v=2*pi*14/T
and solved for T
but this is not the right answer and I'm totally confused as to what I'm doing wrong!

for #2
the only thing I can think of doing is solving for v^2 by doing
155 = v^2/.0144
and then plugging v^2 into
acceleration centripetal = v^2/.0845

The problem here is that the linear velocity v is not the same for both places. (Think of a merry-go-round; the farther you are from the center the faster you are going.)

But what quantity is the same for both places on the rotating disk?

Last edited by a moderator:

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration that an object experiences when it moves in a circular path. It is always directed towards the center of the circle and its magnitude is given by the formula a = v^2/r, where v is the velocity of the object and r is the radius of the circle.

2. How is centripetal acceleration different from normal acceleration?

Centripetal acceleration is a type of acceleration that changes the direction of an object's velocity, while normal acceleration changes the magnitude of an object's velocity. Centripetal acceleration is always perpendicular to the object's velocity, while normal acceleration can be in any direction.

3. Can centripetal acceleration be negative?

Yes, centripetal acceleration can be negative. Negative centripetal acceleration means that the object is accelerating towards the center of the circle in the opposite direction of its motion.

4. What is the relationship between centripetal acceleration and centripetal force?

Centripetal acceleration is directly proportional to centripetal force. This means that as the centripetal force increases, so does the centripetal acceleration, and vice versa. This relationship is described by the equation a = F/m, where F is the centripetal force and m is the mass of the object.

5. How is centripetal acceleration related to angular velocity?

Centripetal acceleration is directly proportional to angular velocity. This means that as the angular velocity increases, so does the centripetal acceleration, and vice versa. This relationship is described by the equation a = ω^2r, where ω is the angular velocity and r is the radius of the circle.

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