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 .
To determine the direction in which the yo-yo will roll, we need to consider the forces acting on it.
First, let's analyze the forces when the yo-yo is rolling to the left. In this case, the force of static friction between the yo-yo and the floor will act to the right. The tension in the string will act at an angle β with respect to the horizontal direction. We know that the static frictional force must be greater than or equal to the force pulling the yo-yo to the left to prevent slippage.
Now, let's consider the forces when the yo-yo is rolling to the right. In this case, the force of static friction between the yo-yo and the floor will act to the left, opposing the motion. Again, the tension in the string will act at an angle β with respect to the horizontal direction. The static frictional force must be greater than or equal to the force pulling the yo-yo to the right to prevent slippage.
Using trigonometry, we can break down the tension force into its horizontal and vertical components. The horizontal component of the tension will be T * sin(β), and the vertical component will be T * cos(β), where T is the tension in the string.
To determine the maximum static frictional force, we use the formula:
F_friction_max = μ * F_normal
where μ is the coefficient of static friction and F_normal is the normal force acting on the yo-yo.
The normal force, in this case, will be the weight of the yo-yo, which is given by:
F_normal = m * g
where m is the mass of the yo-yo and g is the acceleration due to gravity.
Now, we can set up the inequalities for the forces to determine the rolling direction:
For rolling to the left:
T * sin(β) ≤ F_friction_max
Substituting the values, we get:
T * sin(β) ≤ μ * m * g
Rearranging the equation, we get:
sin(β) ≤ μ * (m * g) / T (Equation 1)
For rolling to the right:
T * sin(β) ≥ F_friction_max
Substituting the values, we get:
T * sin(β) ≥ μ * m * g
Rearranging the equation, we get:
sin(β) ≥ μ * (m * g) / T (Equation 2)
Comparing Equations 1 and 2, we can see that the inequality signs will be opposite. Therefore, the correct answer is:
a) Yo-Yo rolls to the left if sin β < R_1/R_2, and to the right if sin β > R_1/R_2.