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An 7.8-cm -diameter, 340 g solid sphere is released from rest at the top of a 1.5-m -long, 16 ∘ incline. it rolls, without slipping, to the bottom. what is the sphere's angular velocity at the bottom of the incline? what fraction of its kinetic energy is rotational?

User Euraad
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Final answer:

The sphere's angular velocity at the bottom of the incline can be found using the conservation of mechanical energy, and the fraction of its kinetic energy that is rotational is 2/5.

Step-by-step explanation:

To calculate the sphere's angular velocity at the bottom of the incline, we will use the conservation of mechanical energy. The total mechanical energy at the top is equal to the potential energy, which is all converted to translational and rotational kinetic energy at the bottom. We will not consider any energy lost as heat due to non-existent friction or air resistance, since rolling without slipping is a perfectly efficient translation to rotation.

The potential energy at the top (PE) is given by:

PE = mgh
Where m is the mass of the sphere, g is the acceleration due to gravity (9.8 m/s2), and h is the height of the incline calculated from the length of the incline (l) and the sine of the angle of the incline (θ).

Translational kinetic energy (KT) at the bottom is given by:
KT = ½ mv2

Rotational kinetic energy (KR) is given by:

KR = ½ Iω2

Where I is the moment of inertia of a solid sphere (¾ mr2), and ω is the angular velocity.

Solving for the total energy at the bottom (Etotal) gives us:
Etotal = KT + KR

Because the sphere rolls without slipping, the linear velocity (v) is related to the angular velocity (ω) by the radius (r):
v = rω

Now we can equate the potential energy at the top to the sum of translational and rotational kinetic energies at the bottom:
mgh = ½ mv2 + ½ (¾ mr2)ω2

Solving for ω using v = rω, we get:
ω = √(10gh / 7r2)

To find the fraction of kinetic energy that is rotational, we use the following ratio:
Fraction = KR / Etotal

Substitute the values of KR and Etotal and simplify to get:
Fraction = (¾ mr2ω2 / 2) / (mv2 / 2 + ¾ mr2ω2 / 2) = ¾ / (1 + ¾ ) = ¾ / ¾ = 2/5

User GPGVM
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