A bead slides without friction around a loopthe-loop.
The bead is released from a height
of 18.9 m from the bottom of the looptheloop which has a radius 7 m.
The acceleration of gravity is 9.8 m/s^2.
What is its speed at point A ?
Answer in units of m/s.
Monica,
By any chance do you have a diagram of the problem? Unfortunately without a diagram we cannot tell where point A is.
r=7 m
h=18.9 m
From the conservation of energy, we have
Ki+Ui=Kf+Uf
0+mgh=mv^2/2 +mg(2R)
v^2=2g(h-2R).
Therefore
v=sqrt(2g(h-2R))
=sqrt(2(9.8m/s^2)[18.9m-2(7m)]
=9.8 m/s
Well, if I were the bead, I would be feeling quite loopy right now! Let's calculate the speed at point A.
To find the speed, we can make use of conservation of energy. At point A, the bead is at the bottom of the loop, so all its potential energy has been converted into kinetic energy.
The potential energy at the initial position is given by mgh, where m is the mass of the bead, g is the acceleration due to gravity, and h is the height from the bottom of the loop.
The kinetic energy at point A is given by 1/2 mv^2, where v is the speed of the bead at point A.
Since energy is conserved, we can equate the two equations:
mgh = 1/2 mv^2
We know the height h is 18.9 m and the radius of the loop is 7 m. Let's assume the mass of the bead is 1 kg (just to keep it simple).
m = 1 kg
g = 9.8 m/s^2
h = 18.9 m
Plugging these values into the equation, we have:
1 kg * 9.8 m/s^2 * 18.9 m = 1/2 * 1 kg * v^2
v^2 = 2 * 9.8 m/s^2 * 18.9 m
v^2 = 353.16 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 18.8 m/s
So, the speed of the bead at point A is approximately 18.8 m/s. Don't loop out of control now!
To find the speed of the bead at point A, we can use the principle of conservation of energy. At point A, the bead has maximum gravitational potential energy and zero kinetic energy.
The gravitational potential energy (PE) at point A is given by:
PE = mgh
Where:
m = mass of the bead (which we can assume to be 1 kg unless specified)
g = acceleration due to gravity (9.8 m/s^2)
h = height from the bottom of the loop to point A (which is equal to the radius of the loop)
In this case, h = 7 m. So, the potential energy at point A can be calculated as:
PE = 1 kg * 9.8 m/s^2 * 7 m = 68.6 Joules
At point A, all the potential energy is transformed into kinetic energy (KE) because there is no friction. We can equate the potential energy to the kinetic energy:
PE = KE
So, the kinetic energy at point A is also 68.6 Joules.
The kinetic energy (KE) is given by the equation:
KE = (1/2) * m * v^2
Where:
m = mass of the bead (again, 1 kg)
v = speed of the bead
Rearranging the equation, we can solve for the speed (v):
v^2 = (2 * KE) / m
Substituting the given values, we get:
v^2 = (2 * 68.6 J) / 1 kg
v^2 = 137.2 m^2/s^2
Taking the square root of both sides, we find:
v = sqrt(137.2) = 11.71 m/s
Therefore, the speed of the bead at point A is approximately 11.71 m/s.