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 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.
To determine in which direction the yo-yo will roll (to the left or to the right), we need to consider the forces acting on the yo-yo.
When the yo-yo starts to roll without slipping, the frictional force between the yo-yo and the floor provides the torque necessary to make it roll. The direction of the frictional force depends on the direction of the applied force, which is determined by the angle β at which the string is pulled.
To analyze the forces, we need to consider the torque about the center of mass of the yo-yo. The torque due to the frictional force is given by:
τ_friction = f_friction * R
where f_friction is the frictional force and R is the radius of the yo-yo.
The frictional force can be expressed as:
f_friction = μ * N
where μ is the coefficient of static friction and N is the normal force.
The normal force N is equal to the weight of the yo-yo:
N = mg
where m is the mass of the yo-yo and g is the acceleration due to gravity.
Substituting the expressions for f_friction and N into the torque equation, we get:
τ_friction = μ * mg * R
For the yo-yo to start rolling without slipping, the torque due to the frictional force must be equal to the torque due to the tension in the string. The torque due to the tension in the string is given by:
τ_tension = T * R1
where T is the tension in the string and R1 is the radius of the inner part of the yo-yo.
Setting these torques equal to each other, we have:
μ * mg * R = T * R1
Simplifying, we find:
T = μ * mg * (R / R1)
The tension in the string is directly proportional to the sine of the angle β, so we can write:
T = k * sin β
where k is a constant.
Substituting this expression for T into the previous equation, we get:
k * sin β = μ * mg * (R / R1)
Now, let's examine the conditions for the yo-yo to roll to the left and to the right:
a) The yo-yo rolls to the left if sin β < R1/R2, and to the right if sin β > R1/R2.
b) The yo-yo rolls to the left if sin β > R1/R2, and to the right if sin β < R1/R2.
c) The yo-yo rolls to the left if cos β < R1/R2, and to the right if cos β > R1/R2.
d) The yo-yo rolls to the left if cos β > R1/R2, and to the right if cos β < R1/R2.
Among the given options, the correct answer is:
d) The yo-yo rolls to the left if cos β > R1/R2, and to the right if cos β < R1/R2.