Students will learn to solve Yo-Yo type problems. That is problems where an object is rotating in free fall yet attached to a string (or some other similar restraint).
angular accelerations are produced by net torques,with inertia opposing acceleration; this is the rotational analog of
the net torque is the vector sum of all the torques acting on the object. When adding torques it is necessary to subtract CW from CCW torques.
individual torques are determined by multiplying the force applied by the perpendicular component of the moment arm
A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the the disk is allowed to fall and the string is held so that it unravels without sliding as the disk is falling, what will be the wooden disk's acceleration?
First we'll draw a free body diagram for the situation. In the diagram below the force of gravity and the tension from the string have been labeled and are drawn from they points they are acting.
Let's start by applying Newton's second law to this object and we'll let the positive direction be downward.
As you can see, we have two unknowns in this equation( and ), so we're going to use Newton's second law for rotation as our second equation and solve for . The rotation axis is at the cm of the cylinder, thus there is no torque from gravity.
Now we will substitute in values for and . In order to calculate the moment of inertia of the wooden disk, we use the known formula for a disk being rotated about it's center.
We can now plug this back into our original Newton's second law equation from the first step.
Watch this Explanation
A game of tug-o-war is played … but with a twist (ha!). Each team has its own rope attached to a merry-go-round. One team pulls clockwise, the other counterclockwise. Each pulls at a different point and with a different force, as shown.
- Who wins?
- By how much? That is, what is the net torque?
- Assume that the merry-go-round is weighted down with a large pile of steel plates. It is so massive that it has a moment of inertia of . What is its angular acceleration?
- How long will it take the merry-go-round to spin around once completely?
A spherical Yo-Yo is given to you for Christmas. Being the good physicist that you are, you immediately calculate it’s acceleration and then test it with a motion detector. The Yo-Yo, of mass of 200g and a radius of 10cm. is a solid sphere (I = 2/5MR2), with two strings attached, both at a 20° from the top. See figure. The string thickness is negligible compared to the Sphere radius.
- Calculate the final linear velocity of the Yo-Yo when dropped from a height of 0.5m.
- Calculate the angular acceleration of the Yo-Yo.
- If the string thickness is not ignored, how would this change your results in parts a. and b.
- a. team b. c. d.
- a. 2.6 m/s b. 72 c. Effectively makes r bigger, thus acceleration goes up and thus final velocity goes up.