Two equal mass balls (one red and the other blue) are dropped from the same height, and rebound off the floor. The red ball rebo
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Answer:
Explanation:
The first thing to have to make a diagram where we indicate the forces acting on the block, see attached. We have the following
The weight (W) that is vertical
The friction force (fr) that opposes the movement
The driving force (F)
The normal (N) which is the reaction to the support of the body and is perpendicular to the surface
We write a reference system where one axis is parallel to the surface (x axis) and the other is perpendicular to it (Y axis) we decompose the forces on these axes, we see that the only force we have to decompose r is the weight
sin θ = Wx / W ⇒ Wx = W sin θ
cos θ = Xy / W ⇒ Wy = Wcos θ
We write Newton's equations for each axis
N- Wy = 0
F -fr -Wx = m a (1)
The expression for the friction force is
fr = μ N
N = Wy
N = W cos θ
fr = μ mg cosT
We hope, we simplify the acceleration of equation 1
a = F - μ mg cos θ -mg sinθ
a = F - mg (μ cos θ -sin θ)
This is the general equation for movement.
Answer:
Explanation:
Length of bar = L
mass of bar = M
mass of each ball = m
Moment of inertia of the bar about its centre perpendicular to its plane is
Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is
Total moment of inertia of the system about the centre of the bar perpendicular to its plane is
I = I1 + I2