Physics, 19.02.2020 00:17 golffuture666

For the following objects, which all roll without slip- ping, determine the rotational kinetic energy about the center of mass as a percentage of the total kinetic energy: (a) a solid sphere, (b) a thin spherical shell, and (c) a tiin cylindrical shell.

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Answer from: kiaraphilman2956

The answers to the question are

(a) 28.57 %

(b) 40 %

Explanation:

The following are the relations between total and rotational kinetic energies of a sphere

$KE_{Total} = KE_{Translational} +KE_{Rotational}$ =

$KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} I\omega ^{2}$

ω = angular velocity

v = velocity

m = mass

I = Moment of inertia = $\frac{2}{5} *m*r^{2}$ and ω = v/r

Therefore $KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} )*m*r^{2}*\frac{v}{r} ^{2}$ = $KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{5} mv^{2} = \frac{7}{10} mv^{2}$

Therefore $\frac{KE_{Total} }{KE_{Rotational} } Percent= \frac{\frac{1}{5} mv^{2} }{\frac{7}{10}mv^{2} } *100 = \frac{10}{35} *100 = 28.57 %$

= 28.57 %

(b) for a thin sherical shell, I = $\frac{2}{3} mr^{2}$ Therefore $KE_{Rotational} = \frac{1}{2} I\omega^{2}$

= $(\frac{1}{2}) (\frac{2}{3}) mr^{2} *\omega ^{2}$ = $(\frac{1}{3}) I*(\frac{v}{r} )^{2}$ = 1/3mv²

KE(Total) = KE(Transitional) + KE(Rotational) = 1/2mv² + 1/3mv² = 5/6mv²

KE(Transitional) /KE(Total) = (1/3mv²)/(5/6mv²) = 6/15

Percentage total = 6*100/15 = 40 %

(c) Thin cylindrical shell = I = $mr^{2}$ and $KE_{Rotational} = \frac{1}{2} I\omega^{2} = \frac{1}{2} mr^{2}(\frac{v}{r}) ^{2} = \frac{1}{2} mv^{2}$

Therefore $KE_{Total} = KE_{Translational} +KE_{Rotational}$

$KE_{Total} =\frac{1}{2} m v^{2} + \frac{1}{2} mv ^{2} =mv ^{2}$ and $\frac{KE_{Rotational}}{KE_{Total} } = \frac{ \frac{1}{2} m v^{2}}{mv ^{2} } =\frac{1}{2}$

In percentage = $\frac{1}{2} * 100$ = 50 %

Answer from: Quest

conduction, convection, radiation

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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For the following objects, which all roll without slip- ping, determine the rotational kinetic energ...