A 190 g basketball has a 37.3 cm diameter and may be approximated as a thin spherical shell.
Starting from rest, how long will it take a basketball to roll without slipping 2.48 m down an incline that makes an angle of 46◦ with the horizontal? The moment of inertia of a thin spherical shell of radius R and mass m is I = 23mR2, the acceleration due to gravity is 9.8 m/s2 , and the coefficient of friction is 0.52 .
Answer in units of s.
To find the time it takes for the basketball to roll without slipping down the inclined plane, we can use the equations of rotational motion and consider the forces acting on the basketball.
First, let's find the net torque acting on the basketball. The net torque is the product of the frictional force applied at the point of contact between the basketball and the incline, and the radius of the basketball. The frictional force can be calculated using the coefficient of friction and the normal force. The normal force acting on the basketball is equal to its weight, which is the product of its mass and the acceleration due to gravity. Thus, we have:
Frictional force (Ff) = coefficient of friction * normal force
Normal force (N) = mass * acceleration due to gravity
Substituting the given values:
mass = 190 g = 0.19 kg (convert to kg)
acceleration due to gravity = 9.8 m/s^2
We have:
N = 0.19 kg * 9.8 m/s^2 = 1.862 N
Now, we can calculate the frictional force:
Ff = 0.52 * 1.862 N = 0.96744 N
Next, let's calculate the net torque by multiplying the frictional force by the radius of the basketball:
Net torque = Ff * radius
Given the diameter of the basketball (37.3 cm), the radius (R) can be calculated by dividing the diameter by 2 and converting it to meters:
radius (R) = (37.3 cm / 2) / 100 = 0.1865 m
Now, we can calculate the net torque:
Net torque = 0.96744 N * 0.1865 m = 0.18032968 N•m
We can now use the torque equation to find the angular acceleration of the basketball:
Torque = moment of inertia * angular acceleration
The moment of inertia of a thin spherical shell is given by the equation:
moment of inertia (I) = 2/3 * mass * radius^2
Substituting the given values:
mass = 0.19 kg
radius = 0.1865 m
We can calculate the moment of inertia:
I = 2/3 * 0.19 kg * (0.1865 m)^2 = 0.01283286 kg•m^2
Now, we can rearrange the torque equation to solve for angular acceleration:
angular acceleration = Torque / moment of inertia
angular acceleration = 0.18032968 N•m / 0.01283286 kg•m^2 = 14.053 rad/s^2
Next, we need to find the angular velocity of the basketball at the bottom of the incline. We can use the equation of rotational motion:
Angular velocity^2 = Initial angular velocity^2 + 2 * angular acceleration * angular displacement
At the top of the incline, the basketball is at rest, so its initial angular velocity is 0. The angular displacement can be calculated using the arc length formula:
Angular displacement = (arc length) / radius
Arc length = distance rolled without slipping = 2.48 m
Substituting the values:
angular displacement = 2.48 m / 0.1865 m = 13.314 rad
Now we can use the rotational motion equation to find the angular velocity:
0^2 = 0 + 2 * 14.053 rad/s^2 * 13.314 rad
0 = 2 * 14.053 rad/s^2 * 13.314 rad
0 = 372.255732 rad^2/s^2
Now, square root 372.255732 to get velocity
angular velocity = √(372.255732) rad/s = 19.301810 rad/s
Finally, we can find the time it takes for the basketball to roll down the incline:
time = angular displacement / angular velocity
time = 13.314 rad / 19.301810 rad/s = 0.689 s
Therefore, it will take approximately 0.689 seconds for the basketball to roll without slipping down the incline.