Question

You are part of an engineering team trying to reproduce this in a mechanical cat. Your prototype, aptly named Katt, when dropped motionless, upside down from 2.5 m, can right itself before it hits the ground by rotating its "tail". Cat’s body is a solid cylinder 30 cm in length and 15 cm in diameter, with a mass of 3.0 kg. Attached to the center of one end of the body is Cat's tail, a 20-cm long rod which extends out perpendicular to Katt's body and has only 10% the mass of the body. The tail will rotate due to a small electric motor in the body that produces a constant torque. In order to order the piece, determine the angular speeds it will need to deal with.

Answer 1

36.26 rad/s

Step-by-step explanation:

From work-kinetic energy principles,

rotational kinetic energy of mechanical cat = potential energy change

¹/₂Iω² = M₁gh where I = rotational inertia of mechanical cat, ω = angular speed, M₁ = mass of mechanical cal = m + M where m = mass of tail and M = mass of cylindrical body, h = height = 2.5 m

I = I₁ + I₂ where I₁ = rotational inertia of tail about cylindrical body = I₃ + mL²/4 (parallel axis theorem). where L = length of tail = 20 cm = 0.2 m

I₃ = rotational inertia of tail = ¹/₁₂mL² and I₂ = rotational inertia of cylindrical body about central axis = ¹/₂MR² where R = radius of cylinder = 15/2 cm = 7.5 cm = 0.075 m

I = I₁ + I₂ = I₃ + mL²/4 + ¹/₂MR² = ¹/₁₂mL² + mL²/4 + ¹/₂MR²

I = ¹/₃mL² + ¹/₂MR² since m = 0.1M,

I = ¹/₃(0.1M)L² + ¹/₂MR²

I = (¹/₃₀L² + ¹/₂R²)M

¹/₂Iω² = M₁gh and M₁ = m + M = 0.1M + M = 1.1M = ¹¹/₁₀M

¹/₂(¹/₃₀L² + ¹/₂R²)Mω² = ¹¹/₁₀Mgh

(¹/₃₀L² + ¹/₂R²)ω² = ¹¹/₅gh

ω = √[¹¹/₅gh/(¹/₃₀L² + ¹/₂R²)]

ω
`\sqrt{((11)/(5)gh )/((1)/(30)L^(2) + (1)/(2)R^(2) ) } \\\sqrt{((11)/(5)X9.8X2.5 )/((1)/(30)0.2^(2) + (1)/(2)0.075^(2) ) }\\\sqrt{(5.39)/(0.0013 + 0.0028) }\\ \sqrt{(5.39)/(0.0041) }\\√(1314.6342) \\= 36.26 rad/s`