Respuesta :
Explanation:
Given that,
(1). Mass of solid cylinder A= 100 g
Radius = 10 cm
Mass of solid cylinder B= 200 g
Radius = 10 cm
(2). Mass of solid cylinder A= 100 g
Radius = 10 cm
Mass of solid cylinder B= 100 g
Radius = 20 cm
We need to calculate the moment of inertia of solid cylinder A and B
Using formula of moment of inertia
[tex]I=\dfrac{1}{2}mR^2[/tex]
But we know that,
For pure rolling rotation,
An object under the influence of gravity does not depend on the object's mass and its size.
Therefore, Both solid cylinders reach on same time.
(3). Mass of solid cylinder A= 100 g
Radius = 10 cm
Mass of hollow cylinder B= 100 g
Radius = 10 cm
We need to calculate the moment of inertia of solid cylinder A
Using formula of moment of inertia
[tex]I=\dfrac{1}{2}mR^2[/tex]
[tex]I_{A}=\dfrac{1}{2}\times100\times10^{-3}\times(10\times10^{-2})^2[/tex]
[tex]I_{A}=0.0005\ kg-m^2[/tex]
We need to calculate the moment of inertia of hollow cylinder B
Using formula of moment of inertia
[tex]I=mR^2[/tex]
[tex]I_{B}=100\times10^{-3}\times(10\times10^{-2})^2[/tex]
[tex]I_{B}=0.001\ kg-m^2[/tex]
Here, The moment of inertia of A is less for hollow cylinder.
Therefore, Solid cylinder will reach first.
(4). Mass of solid cylinder A= 100 g
Radius = 10 cm
Mass of solid sphere B= 100 g
Radius = 10 cm
We need to calculate the moment of inertia of solid cylinder A
Using formula of moment of inertia
[tex]I=\dfrac{1}{2}mR^2[/tex]
[tex]I_{A}=\dfrac{1}{2}\times100\times10^{-3}\times(10\times10^{-2})^2[/tex]
[tex]I_{A}=0.0005\ kg-m^2[/tex]
We need to calculate the moment of inertia of solid sphere B
Using formula of moment of inertia
[tex]I=\dfrac{2}{5}mR^2[/tex]
[tex]I_{B}=\dfrac{2}{5}\times100\times10^{-3}\times(10\times10^{-2})^2[/tex]
[tex]I_{B}=0.0004\ kg-m^2[/tex]
Here, The moment of inertia of A is more for solid sphere.
Therefore, Solid sphere will reach first
Hence, This is the required solution.
Given the mass, radius and type of solid, we have;
- The race is a tie
- The race is a tie
- The hollow cylinder will win the race
- The sphere will win the race
How can how the race outcome be known?
1. Mass of the solid cylinder, MA = 100 g = 0.1 kg.
Radius of the solid cylinder, RA = 10 cm = 0.1 m
Mass of the solid cylinder, MB = 200 g = 0.1 kg
[tex]a[/tex]
Radius of the solid cylinder, RB = 10 cm
Which gives;
[tex]m \times g \times h = \frac{1}{2} \times m \times {v}^{2} + \frac{1}{2} \times i \times { \omega}^{2} [/tex]
Which gives;
[tex]mgh = \mathbf{\frac{1}{2} m {v}^{2} + \frac{1}{2} m {r}^{2} { \omega}^{2} [/tex]
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{2} m {r}^{2} { \frac{v}{r} }^{2} [/tex]
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{2} m {v}^{2} [/tex]
[tex]mgh = m {v}^{2}[/tex]
[tex]gh = \mathbf{{v}^{2}}[/tex]
Therefore, the speed depends on the height, such that both cylinders, have the same speed when released from the same height.
- The race is a tie
2. Given that the speed of a solid cylinder depends only on the height, we have;
- The race is a tie
3. For the solid cylinder, we have;
[tex]mgh = \mathbf{\frac{1}{2} m {v}^{2} + \frac{1}{2} m {r}^{2} { \frac{v}{r} }^{2} } [/tex]
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{2} m{v}^{2} [/tex]
[tex]mgh = m{v}^{2} [/tex]
[tex]gh = {v}^{2} [/tex]
For the hollow cylinder, we have;
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{4} m {r}^{2} { \frac{v}{r} }^{2} [/tex]
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{4} m{v}^{2} [/tex]
[tex]mgh = \frac{3}{4} m{v}^{2} [/tex]
[tex]gh = \frac{3}{4} {v}^{2} [/tex]
Therefore, the velocity of the hollow cylinder, is higher than that of the solid cylinder.
- The hollow cylinder will win the race.
4. For the solid cylinder, we have;
[tex]gh = \frac{3}{4} {v}^{2} [/tex]
For the sphere, we have;
[tex]mgh = \frac{1}{2} m {v}^{2} + \frac{1}{5} m{v}^{2} [/tex]
[tex]gh = \frac{7}{10} {v}^{2} (sphere)
Therefore, the sphere will win the race.
Learn more about moment of inertia here:
