Question
A regulation basketball has a 14 cm diameter and may be approximated as a thin spherical shell.
How long will it take a basketball starting from rest to roll without slipping 2.8 m down an incline that makes an angle of 41.1° with the horizontal? The acceleration of gravity is 9.81 m/52
Answer in units of s.
How long will it take a basketball starting from rest to roll without slipping 2.8 m down an incline that makes an angle of 41.1° with the horizontal? The acceleration of gravity is 9.81 m/52
Answer in units of s.
Answers
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Answered by
GPT 3.5
To solve this problem, we'll use the concept of rotational motion.
First, let's find the mass of the basketball. The volume of a sphere is given by V = (4/3)πr^3, where r is the radius. The diameter of the basketball is 14 cm, so the radius is 7 cm = 0.07 m. Plugging this into the volume formula, we get:
V = (4/3)π(0.07)^3 = 0.015292 m^3
The density of a basketball is about 600 kg/m^3, so the mass of the basketball is:
m = density * volume = 600 kg/m^3 * 0.015292 m^3 = 9.1752 kg
Now, let's find the moment of inertia of the basketball. The moment of inertia I for a thin spherical shell is given by I = (2/3)mr^2, where m is the mass and r is the radius. Plugging in the values, we get:
I = (2/3) * 9.1752 kg * (0.07 m)^2 = 0.0467 kg⋅m^2
Next, we'll use the torque equation τ = Iα to relate the angular acceleration α with the applied torque τ. Since the basketball is assumed to roll without slipping, α and a (linear acceleration) are related by α = a/r, where r is the radius. Plugging in the values, we get:
τ = I * (a/r) = (2/3) * 9.1752 kg * (0.07 m)^2 * (a/0.07 m) = 0.0975 a kg⋅m^2
When the basketball starts rolling without slipping, the torque acting on it is due to the component of the gravitational force along the incline, which is mgsin(θ), where g is the acceleration due to gravity and θ is the angle of the incline. Plugging in the values, we get:
τ = mgsin(θ) = 9.1752 kg * 9.81 m/s^2 * sin(41.1°) = 58.2587 N⋅m
Setting this equal to the torque from the angular acceleration, we get:
0.0975 a = 58.2587
Solving for a, we find a = 596.9745 m/s^2.
Now, let's use the kinematic equation s = ut + (1/2)at^2 to find the time t it takes for the basketball to roll 2.8 m down the incline. Since the basketball starts from rest, the initial velocity u is 0. Plugging in the values, we get:
2.8 m = (1/2) * 596.9745 m/s^2 * t^2
Simplifying, we find:
t^2 = (2 * 2.8 m) / 596.9745 m/s^2 = 0.0093462 s^2
Taking the square root of both sides, we get:
t = 0.0966546 s
Therefore, it will take approximately 0.097 s for the basketball to roll without slipping 2.8 m down the incline.
First, let's find the mass of the basketball. The volume of a sphere is given by V = (4/3)πr^3, where r is the radius. The diameter of the basketball is 14 cm, so the radius is 7 cm = 0.07 m. Plugging this into the volume formula, we get:
V = (4/3)π(0.07)^3 = 0.015292 m^3
The density of a basketball is about 600 kg/m^3, so the mass of the basketball is:
m = density * volume = 600 kg/m^3 * 0.015292 m^3 = 9.1752 kg
Now, let's find the moment of inertia of the basketball. The moment of inertia I for a thin spherical shell is given by I = (2/3)mr^2, where m is the mass and r is the radius. Plugging in the values, we get:
I = (2/3) * 9.1752 kg * (0.07 m)^2 = 0.0467 kg⋅m^2
Next, we'll use the torque equation τ = Iα to relate the angular acceleration α with the applied torque τ. Since the basketball is assumed to roll without slipping, α and a (linear acceleration) are related by α = a/r, where r is the radius. Plugging in the values, we get:
τ = I * (a/r) = (2/3) * 9.1752 kg * (0.07 m)^2 * (a/0.07 m) = 0.0975 a kg⋅m^2
When the basketball starts rolling without slipping, the torque acting on it is due to the component of the gravitational force along the incline, which is mgsin(θ), where g is the acceleration due to gravity and θ is the angle of the incline. Plugging in the values, we get:
τ = mgsin(θ) = 9.1752 kg * 9.81 m/s^2 * sin(41.1°) = 58.2587 N⋅m
Setting this equal to the torque from the angular acceleration, we get:
0.0975 a = 58.2587
Solving for a, we find a = 596.9745 m/s^2.
Now, let's use the kinematic equation s = ut + (1/2)at^2 to find the time t it takes for the basketball to roll 2.8 m down the incline. Since the basketball starts from rest, the initial velocity u is 0. Plugging in the values, we get:
2.8 m = (1/2) * 596.9745 m/s^2 * t^2
Simplifying, we find:
t^2 = (2 * 2.8 m) / 596.9745 m/s^2 = 0.0093462 s^2
Taking the square root of both sides, we get:
t = 0.0966546 s
Therefore, it will take approximately 0.097 s for the basketball to roll without slipping 2.8 m down the incline.
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