A uniform 5.32 kg solid sphere of radius 0.344 m that is free to roll on a horizontal surface has a light axle passing through its diameter as shown. The axle is connected to a like yoke and in turn a light rope as shown. A constant tension of 7.81 N is applied ot the rope and the sphere starts to roll without slipping on the rough surface.
a) What is the angular acceleration of the sphere in rad/s2 ? (Hint: Find the acceleration of the sphere's center of mass first.)
b) What is the magnitude of the static frictional force that acts on the sphere in Newtons? (ìs is NOT needed for this part of the question.)
c) If the coefficient of static friction is 0.211, what is the maximum tension in Newtons that could be applied to the rope for which the sphere would still roll without slipping? (i.e. If the tension is greater than this value, the sphere would start to slip.)
In order to find the angular acceleration of the sphere, we need to first find its linear acceleration. We can do this by considering the forces acting on the sphere.
Let's break down the forces:
1. Gravitational force (weight): The weight of the sphere can be calculated using the formula F = mg, where m is the mass of the sphere and g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, the weight is 5.32 kg multiplied by 9.8 m/s^2.
2. Tension force: The constant tension force applied to the rope is 7.81 N. This force will cause the sphere to move and accelerate.
a) To find the angular acceleration, we need to find the linear acceleration first.
The torque produced by the tension force is responsible for the angular acceleration of the sphere. The torque is given by the formula τ = Iα, where τ is the torque, I is the moment of inertia of the sphere, and α is the angular acceleration.
For a solid sphere rolling without slipping, the moment of inertia is given by I = (2/5) * m * r^2, where m is the mass of the sphere and r is the radius of the sphere.
Now, we can equate the torque due to the tension force to the torque due to the frictional force:
τ = Fr,
where F is the force of friction and r is the radius of the sphere.
Since the sphere is just at the point of rolling without slipping, the frictional force can be related to the normal force (N) and the coefficient of static friction (μs) using the equation F = μs * N.
The normal force is equal to the weight of the sphere in this case, so the frictional force is μs * mg.
b) Now that we have the expression for the frictional force, we can find its magnitude. Plug in the given coefficient of static friction (0.211), the mass of the sphere (5.32 kg), and the acceleration due to gravity (9.8 m/s^2) into the equation and calculate the frictional force.
c) To find the maximum tension in the rope for which the sphere will roll without slipping, we need to equate the torque due to the tension force to the torque due to the maximum static frictional force.
τ = Fr_max,
where r_max is the distance between the center of mass of the sphere and the point where the tension force is applied. Solve this equation for r_max and calculate the value.
Note that in order to solve this problem, we have made assumptions that the sphere is rolling without slipping and there is no air resistance.