Two questions about centripetal acceleration,

[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

what am I doing wrong? Please help!

 

[alphysicist]

Hi 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

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?

what am I doing wrong? Please help!