A hollow 0.358 kg sphere rolls without slipping down an inclined plane that makes an angle of 41.0o with the horizontal direction. The sphere is released from rest a distance 0.734 m from the lower end of the plane.
a. How fast is the hollow sphere moving as it reaches the end of the plane?
b. At the bottom of the incline, what fraction of the total kinetic energy of the hollow sphere is rotational kinetic energy?
This is what i got but i'm not sure its correct. Any help would be much appreciated!
a)
cons. of energy
1/2mv^2+1/2Iw^2=mgLsinθ
where
I=2/3mr^2
w=v/r
then
1/2mv^2+1/3mv^2=mgLsinθ
5/6mv^2=mgLsinθ
v=(6gLsinθ/5)^1/2
v=(6*9.8*0.734*sin41/5)^1/2 = 2.38 m/s
b)
K=1/2mv^2+1/3mv^2
K=5/6mv^2
Krotational=1/3mv^2
Krot/K=2/5=0.4
What you did makes sense to me. I guess we are assuming that the sphere has a very thin wall thickness to get the (2/3) m r^2
To solve this problem, we can use the principles of energy conservation and rotational motion. Let's break it down step by step:
a) To find the speed of the hollow sphere as it reaches the end of the plane, we can apply the principle of conservation of energy. At the start, the sphere has gravitational potential energy, which is converted into kinetic energy as it moves down the inclined plane.
Using the conservation of energy equation:
1/2mv^2 + 1/2Iw^2 = mgLsinθ
Where:
m = mass of the hollow sphere (0.358 kg)
v = speed of the sphere
I = moment of inertia of the hollow sphere (2/3mr^2)
w = angular velocity of the sphere (v/r)
g = acceleration due to gravity (9.8 m/s^2)
L = distance the sphere travels along the inclined plane (0.734 m)
θ = angle of the inclined plane (41 degrees)
First, we need to find the angular velocity (w) in terms of speed (v) and radius (r). Since the sphere is rolling without slipping, the speed of the sphere is equal to the product of its angular velocity (w) and radius (r). Therefore:
w = v / r
Next, we substitute the value of the moment of inertia (I) and the angular velocity (w) into the conservation of energy equation:
1/2mv^2 + 1/2(2/3mr^2)((v/r)^2) = mgLsinθ
Simplifying the equation:
1/2mv^2 + 1/3mv^2 = mgLsinθ
Combining like terms:
5/6mv^2 = mgLsinθ
Finally, we solve for v:
v = sqrt((6gLsinθ) / 5)
Plug in the values:
v = sqrt((6 * 9.8 * 0.734 * sin(41)) / 5)
v ≈ 2.38 m/s
Therefore, the hollow sphere is moving at approximately 2.38 m/s as it reaches the end of the plane.
b) To find the fraction of the total kinetic energy that is rotational kinetic energy at the bottom of the incline, we compare the rotational kinetic energy (Krotational) and the total kinetic energy (K).
Total kinetic energy of the sphere is given by:
K = 1/2mv^2 + 1/2Iw^2
Substituting the value of the moment of inertia (I) and the angular velocity (w):
K = 1/2mv^2 + 1/2(2/3mr^2)((v/r)^2)
Simplifying the equation:
K = 5/6mv^2
Rotational kinetic energy is given by:
Krotational = 1/2Iw^2
Substituting the value of the moment of inertia (I) and the angular velocity (w):
Krotational = 1/2(2/3mr^2)((v/r)^2)
Simplifying the equation:
Krotational = 1/3mv^2
To find the fraction of the total kinetic energy that is rotational kinetic energy, we calculate:
Krotational / K = (1/3mv^2) / (5/6mv^2)
Simplifying the equation:
Krotational / K = 2/5
Therefore, at the bottom of the incline, the fraction of the total kinetic energy of the hollow sphere that is rotational kinetic energy is 2/5 or 0.4.