If the ball is released from rest at a height of 0.64 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track?Assume the ball is a solid sphere of radius 4.0 cm and mass 0.14 kg
Applying law of conservation of energy:
P.E at height h = sum of translational and rotational K.E
mgh=1/2mv^2 +1/2Iw^2
but v=wr
I =moment of inertia.
=2/5mr^2
i hope u have get the concept Sarah
To determine the angular speed of the ball when it is on the frictionless side of the track, we can make use of the principle of conservation of mechanical energy. The mechanical energy of an object refers to the sum of its potential energy and kinetic energy.
In this case, when the ball is released from rest on the no-slip side of the track, it has gravitational potential energy, which is given by the formula:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height from which the ball is released.
We are given that the height h is 0.64 m, the mass m is 0.14 kg, and the radius r (which will be needed later) is 4.0 cm (or 0.04 m).
Next, let's consider the kinetic energy of the ball when it reaches the frictionless side of the track. At that point, the ball has lost all of its potential energy and has converted it into kinetic energy.
The formula for kinetic energy is:
KE = (1/2) I ω^2
where I is the moment of inertia of the ball and ω is its angular speed.
The moment of inertia of a solid sphere rotating about an axis passing through its center is given by:
I = (2/5) m r^2
Using the given values of mass and radius, we can calculate the moment of inertia.
Now, equating the potential energy (PE) to the kinetic energy (KE) and solving for ω, we can find the angular speed.
mgh = (1/2) I ω^2
Substituting the expressions for PE and KE:
mgh = (1/2) (2/5) m r^2 ω^2
Canceling out the mass (m) on both sides, we get:
gh = (1/5) r^2 ω^2
Simplifying further:
ω^2 = (5gh) / r^2
Taking the square root of both sides:
ω = √(5gh) / r
Now we can substitute the given values and calculate the angular speed of the ball:
ω = √(5 * 9.8 * 0.64) / 0.04
Simplifying the expression:
ω = √(3.136) / 0.04
ω = √78.4
ω ≈ 8.85 rad/s
Therefore, the angular speed of the ball when it is on the frictionless side of the track is approximately 8.85 rad/s.