A yo-yo of mass m rests on the floor (the static friction coefficient with the floor is mu ). The inner (shaded) portion of the yo-yo has a radius R-1 , the two outer disks have radii R-2 . A string is wrapped around the inner part. Someone pulls on the string at an angle Beta (see sketch). The "pull" is very gentle, and is carefully increased until the yo-yo starts to roll without slipping.
For what angles of beta will the yo-yo roll to the left and for what angles to the right?
a) Yo-Yo rolls to the left if sin beta<R-1/R_2 , and to the right if sin beta> R_1/R_2.
b)Yo-Yo rolls to the left if sin beta>R-1/R_2 , and to the right if sin beta<R_1/R_2
c)yo- yo rolls to the left if cos beta< R_1/R_2 and to the right if cos beta>R_1/R-2
d)yo yo rolls to the left if cosbeta>R_1/R_2 and to the right if cosbeta< R_1/R_2 .
The reason I am not answering is that without a drawing I find the question incomprehensible.
Hi Damon,
R_1 is inner radius and R_2 is outer radius of yo yo which looks like concentric circle and beta is the angle in right side of that yo yo. I hope this will help to complete the question. I don't know how to draw image here.
Hi, Damon I got it. Thank you
To determine whether the yo-yo will roll to the left or to the right, we need to consider the forces acting on it.
When the yo-yo starts to roll without slipping, the static friction between the yo-yo and the floor provides the necessary torque to start the rotation. The static friction acts in the opposite direction of the rotation, preventing the yo-yo from slipping.
Let's consider the forces acting on the yo-yo when it starts to roll. There are two main forces: the tension in the string and the static friction.
1. Tension in the string:
The tension force in the string can be broken down into two components: one perpendicular to the radius of the yo-yo and one tangential to the radius. The perpendicular component provides the necessary centripetal force for circular motion. The tangential component produces a torque that opposes the rotation.
2. Static friction:
The static friction acts on the yo-yo in the opposite direction of the rotation. It provides the necessary torque to start the rotation without slipping.
Now, let's analyze the forces in the different scenarios:
For the yo-yo to roll to the left:
The torque provided by the static friction needs to be greater than the torque produced by the tension. This means the static friction force needs to be larger than the tangential component of the tension force.
For the yo-yo to roll to the right:
The torque provided by the static friction needs to be smaller than the torque produced by the tension. This means the static friction force needs to be smaller than the tangential component of the tension force.
To compare the magnitudes of these forces, we can consider the ratios of the forces:
sin(beta) = (tangential component of tension force) / (tension force)
The tangential component of the tension force can be calculated as R-1 * tension, where R-1 is the radius of the inner part of the yo-yo and tension is the tension force.
If the static friction provides enough torque to counteract the torque produced by the tension, then the yo-yo will roll to the left. This is true when sin(beta) < R-1 / R-2.
If the static friction does not provide enough torque to counteract the torque produced by the tension, then the yo-yo will roll to the right. This is true when sin(beta) > R-1 / R-2.
Therefore, the correct answer is:
a) Yo-Yo rolls to the left if sin beta < R-1 / R_2, and to the right if sin beta > R_1 / R_2.