How Accurate Is the Calculation of Angular Acceleration and Time for a Yo-yo?

[ac7597]

Homework Statement

Joe has a simple yoyo, which consists of two disks, each of mass M=0.098 kg and radius R=0.144 meters, connected by a cylindrical spindle of mass m=0.016 kg and radius r=0.01 meters. Joe wraps some string around the spindle; it is made of rough fiber, so it will not slip as it winds and unwinds.

Joe places the yoyo horizontally on a workbench, as shown. He inserts a pin into a tiny hole in center of the bottom disk, so that the entire yoyo is free to rotate without friction around the pin.

What is the moment of inertia of the yoyo around the pin?

Fred now pulls on the string with a constant force T=5.2 Newtons. As he pulls, the string unwinds from the spindle and the yoyo starts to spin.

What is the torque exerted by the string around the pin?

What is the angular acceleration of the yoyo?

How long will it take Fred to pull the string a distance L=0.65 meters?

Relevant Equations

torque=|r|*|F|*sin(theta)= I * angular acceleration

total moment of inertia= (1/2) (0.098kg) ( 0.144m)^2 + (1/2) (0.098kg) ( 0.144m)^2 + (1/2) (0.016kg) (0.01m)^2
total moment of inertia= 2.03*10^(-3) kgm^2

torque= 5.2N * (0.144m) = 0.75N*m
thus: 0.75N*m= 2.03*10^(-3) kgm^2 * angular acceleration
angular acceleration = 368.8 rad/s^2

Is the work correct so far?

 

[tnich]

Looks like you have the string wrapped around the edge of a disk.

 

[ac7597]

Is the torque just involves the cylindrical spindle thus :
torque= (1/2) (0.016kg) (0.01m)^2 ?

 

[haruspex]

Is the torque just involves the cylindrical spindle thus :
torque= (1/2) (0.016kg) (0.01m)^2 ?

Did you mean that? You seem confused between torque and moment of inertia.

 

[tnich]

Is the torque just involves the cylindrical spindle thus :
torque= (1/2) (0.016kg) (0.01m)^2 ?

That wasn't what I meant. I meant that you used the wrong lever arm in computing the torque. I think it would help you to see how to do it if you drew a diagram looking at the yoyo along the direction of the pin, showing the string and the outline of the disks and the spindle.

 

[ac7597]

sorry about the mistake.
total moment of inertia= 2.03*10^(-3) kgm^2

torque= 5.2 * 0.01 = 5.2 E-2
angular acceleration = (5.2 E-2)/ (2.03E-3)= 25.6 rad/s^2

I don't know how to solve for the time.

 

[haruspex]

how to solve for the time.

Through what angle will it have rotated?

 

[ac7597]

90degrees

 

[haruspex]

90degrees

How do you get that?

 

[ac7597]

linear distance = radians * radius
0.65m= radians * 0.01m
radians= 65 rad

thus with kinematics:
65rad= (1/2) (25.6) *time^2
time=2.25 seconds

 

[haruspex]

linear distance = radians * radius
0.65m= radians * 0.01m
radians= 65 rad

thus with kinematics:
65rad= (1/2) (25.6) *time^2
time=2.25 seconds

Looks good.