Answer:
K.E = 0.0075 J
Explanation:
Given data:
Mass of the ball = 1.5 kg
radius, r = 50 cm = 0.5 m
Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec
Now,
the kinetic energy is given as:
K.E =
where,
I is the moment of inertia = mr²
on substituting the values, we get
or
K.E = 0.0075 J
Answer:
The speed at which the ball rolled off the end of the table is 3.3 m/s
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.
The position vector of the ball can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity.
When the ball reaches the ground, its position will be:
r final = (3, -4)
Then:
3 = x0 + v0x · t
-4 = y0 + v0y · t + 1/2 · g · t²
Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:
3 m = v0x · t
-4 m = 1/2 · g · t²
We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:
-4 m = 1/2 · g · t²
-4 m = -1/2 · 9.8 m/s² · t²
8 m / 9.8 m/s² = t²
t = 0.9 s
Then:
3 m = v0x · 0.9s
3 m/ 0.9 s = v0x
v0x = 3.3 m/s
The speed at which the ball roll off the end of the table is 3.3 m/s
