Consider a ball rolling around in a circular path on the inner surface of a cone. the weight of the ball is shown by the vector
1 answer:
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Refer to the diagram shown below.
Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.
The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)
Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N
The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N
If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²
Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.
Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.
The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)
Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N
The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N
If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²
Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.