Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>
Explanation:
- Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
- A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
- You can indeed have a collider with no rigidbody. If there's no rigidbody then Unity assumes the object is static, non-moving.
- If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.
The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N.
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
Answer:
Explanation:
Using the principle of moment, assuming the rod is uniform rod of mass 1 kg
the center of mass of the rod will be at 1 m
assuming the system is in equilibrium,
clockwise moment = anticlockwise moment
let the distance of the man shoulder be x from the center of gravity and also is the pivot point
total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and
2kg + 5 kg = 7 kg for front bucket
( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)
5 + 5 x + x = 7 - 7x
5 + 6x = 7 - 7x
6x + 7x = 7 - 5
13x = 2
x = 2 / 13 = 0.154 m
the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).
