How does acceleration of an object change in relation to its mass
1 answer:
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Answer:
The hoop
Explanation:
We need to define the moment of inertia of the different objects, that is,
DISK:
HOOP:
SOLID SPHERE:
HOLLOW SPHERE
If we have the same acceleration for a Torque applied, then
The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration
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.