Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2. The kinetic energy of ball 2 is 37.0 J. Part A What is the kinetic energy of ball 1?
Express your answer with the appropriate units.
K7 = Value Units

The kinetic energy of the ball 1 is 2.06 J

Explanation:

The kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I is its rotational inertia, ω its angular speed, m its mass and v = its velocity of center of mass.

Let m₁, I₁, v₁, d₁ represent the mass, rotational inertia, speed and diameter of  solid ball 1. and Let m₂, I₂, v₂, d₂ represent the mass, rotational inertia, speed and diameter of  solid ball 2.

Since both objects are spheres, I =2/5mr²

Let r₁ = radius of ball 1 and r₂ = radius of ball 2. Since d₂ = 2d₁

⇒ 2r₂ = 4r₁ ⇒ r₂ = 2r₁

Now, the the kinetic energy of sphere 1 is

K₁ = 1/2I₁ω₁² + 1/2m₁v₁²  ω₁ = v₁/r₁ which is the angular speed of solid ball 1.

K₁ = 1/2(2/5mr²)v₁²/r₁² + 1/2m₁v₁²

K₁ = 1/5m₁v₁² + 1/2m₁v₁²

K₁ = 7/10m₁v₁²

Also, the the kinetic energy of sphere 2 is

K₂ = 1/2I₂ω₂² + 1/2m₂v₂²  ω₂ = v₂/r₂ which is the angular speed of solid ball 2.

K₂ = 1/2(2/5m₂r₂²)v₂²/r₂² + 1/2m₂v₂²

K₂ = 1/5m₂v₂² + 1/2m₂v₂²

K₂ = 7/10m₂v₂²

Now, m₁ = m₂/2 and v₁ = v₂/3

Substituting these into K₁, we have

K₁ = 7/10(m₂/2)(v₂/3)²

K₁ = 7/10 × 1/18m₂v₂²

K₁ = (1/18)(7/10m₂v₂²)

K₁ = K₂/18

K₂ = 37.0 J/18

K₂ = 2.06 J

So, the kinetic energy of the ball 1 is 2.06 J

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the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.