A sphere with a moment of inertia 0.658mr2 and mass m and radius r rolls without slipping along the track shown on a planet where the acceleration due to gravity is g. It starts from rest at a height h = 4.87R above the bottom of the circular loop of radius R.
(For simplicity, neglect the size of the sphere relative to the radius of the loop R and the height h, i.e. r << R, h. Also, while static friction is responsible for the rolling motion, it does no work since it acts through no displacement.)
a) What is the magnitude of the normal force that acts on the sphere at the top of the circular loop as a fraction or multiple of the sphere's weight mg?
To determine the magnitude of the normal force that acts on the sphere at the top of the circular loop, we can consider two forces: the weight of the sphere and the normal force.
At the top of the circular loop, the direction of the normal force is downward, opposite to the direction of the weight. Since the sphere is in equilibrium, the sum of the forces in the vertical direction must be zero.
Let's denote the normal force as N.
The weight of the sphere mg can be expressed as the product of the mass m and the acceleration due to gravity g: mg = m * g.
The net force in the vertical direction is the difference between the weight and the normal force: N - mg.
Since the sphere is in equilibrium, the net force must be zero: N - mg = 0.
Therefore, the magnitude of the normal force is equal to the magnitude of the weight of the sphere: N = mg.
So, the magnitude of the normal force that acts on the sphere at the top of the circular loop is equal to the sphere's weight mg.
Therefore, the magnitude of the normal force that acts on the sphere at the top of the circular loop is equal to the sphere's weight mg.
To find the magnitude of the normal force that acts on the sphere at the top of the circular loop, we need to consider the forces acting on the sphere at that point.
There are two main forces at play: the weight of the sphere and the normal force.
1. Weight: The weight of the sphere, mg, acts vertically downward. Since the acceleration due to gravity on the planet is g, the magnitude of the weight is mg.
2. Normal force: The normal force acts perpendicular to the surface and balances the weight of the sphere.
At the top of the circular loop, the net force acting on the sphere should be directed inward to keep it in circular motion. This inward net force is provided by the normal force.
To determine the magnitude of the normal force, we need to consider the net force and the centripetal force acting on the sphere.
Net force: The net force can be calculated using Newton's second law:
Net force = mass x acceleration
Since the sphere is rolling without slipping, the acceleration can be determined using the relationship between linear acceleration (a) and angular acceleration (α) for a rolling sphere:
a = α x r
where r is the radius of the sphere.
The angular acceleration of a rolling sphere can be given by:
α = Net torque / moment of inertia
Since the sphere is rolling without slipping, the torque causing angular acceleration is the weight of the sphere acting perpendicular to the surface of the track. Therefore, the torque can be calculated as:
Torque = mg x R, where R is the radius of the loop.
Using these values, we can determine the angular acceleration:
α = (mg x R) / moment of inertia
Substituting this value of α into the equation for linear acceleration, we get:
a = ((mg x R) / moment of inertia) x r
Now, the net force acting on the sphere is the centripetal force required to keep it in circular motion. The centripetal force can be calculated as:
Centripetal force = mass x acceleration
Centripetal force = m x a
Substituting the value of a we already calculated, we get:
Centripetal force = m x ((mg x R) / moment of inertia) x r
The net inward force acting on the sphere at the top of the circular loop is provided by the normal force, and it can be equated to the centripetal force:
Normal force - mg = m x ((mg x R) / moment of inertia) x r
Simplifying the equation above, we can solve for the magnitude of the normal force:
Normal force = mg + m x ((mg x R) / moment of inertia) x r
To express the normal force as a fraction or multiple of the sphere's weight (mg), we can divide both sides by mg:
Normal force / mg = 1 + (mg x R) / (moment of inertia x r)
So, the magnitude of the normal force that acts on the sphere at the top of the circular loop is given by the expression:
Normal force / mg = 1 + (mg x R) / (moment of inertia x r)