Physics, 14.04.2020 18:12 Bringuplana

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 kg directly onto the middle of the wheel. The clay is in the shape of a pancake and has a radius of 0.19 m. Assume both the wheel and the chunk of clay can be modeled as solid cylinders (I = ½ MR2 ). What is the new tangential velocity of the wheel and the clay?

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Answer from: historyfanboy101

The tangential velocity of the wheel is 1.79 m/s and the tangential velocity of the clay is 0.97 m/s

From the law of conservation of angular momentum, initial angular momentum = final angular momentum

L = L' + L'' where L = initial angular momentum of wheel, L' = final angular momentum of wheel, and L'' = angular momentum of clay.

Iω = Iω' + I'ω' where

I = rotational inertia of wheel = 1/2MR² where M = mass of wheel = 4.2 kg and R = radius of wheel = 0.35 m, ω = initial angular speed of wheel = 0.98 rev/s = 0.98 × 2π rad/s = 6.16 rad/s, I' = rotational inertia of clay = 1/2mr² where m = mass of clay = 2.9 kg and r = radius of clay = 0.19 m, ω = final angular speed of wheel + clayAngular speed of wheel + clay

So, making ω' subject of the formula, we have

ω' = Iω/(I + I')

ω' = 1/2MR²ω/(1/2MR² + 1/2mr²)

ω' = MR²ω/(MR² + mr²)

Substituting the values of the variables into the equation, we have

ω' = 4.2 kg × (0.35 m)² × 6.16 rad/s/(4.2 kg × (0.35 m)² + 2.9 kg × (0.19 m)² )

ω' = 4.2 kg × 0.1225 m² × 6.16 rad/s/(4.2 kg × 0.1225 m² + 2.9 kg × 0.0361 m² )

ω' = 0.5145 kgm² × 6.16 rad/s/(0.5145 kg m² + 0.1047 kgm)²

ω' = 3.169kgm²rad/s ÷0.6192kgm²

ω' = 5.12 rad/s

The tangential velocity of the wheel

Since tangential speed v = Rω' where

R = radius of wheel = 0.35 m and ω' = angular speed = 5.12 rad/s

So, substituting these into the equation, we have

v = Rω'

v = 0.35 m × 5.12 rad/s

v = 1.792 m/s

v ≅ 1.79 m/s

The tangential velocity of the wheel is 1.79 m/s

The tangential velocity of the clay

Since tangential velocity v' = rω' where

r = radius of clay = 0.19 m and ω' = angular speed = 5.12 rad/s

So, substituting these into the equation, we have

v = rω'

v = 0.19 m × 5.12 rad/s

v = 0.973 m/s

v' ≅ 0.97 m/s

The tangential velocity of the clay is 0.97 m/s

The tangential velocity of the wheel is 1.79 m/s and the tangential velocity of the clay is 0.97 m/s

Learn more about tangential velocity here:

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Answer from: ArmyAllen00

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$

$\omega_{f} \approx 5.116\,\frac{rad}{s}$

The tangential velocities of the wheel and the clay are, respectively:

$v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )$

$v_{f,w} = 1.791\,\frac{m}{s}$

$v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )$

$v_{f,c} = 0.972\,\frac{m}{s}$

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concave lenses are used here. they spread the light out before it reaches the convex lens in the eye, therefore letting the image focus directly on the retina. hyperopia is known to you probably as farsightedness.

Answer from: Quest

this means that the temperature of both substances will be equal

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A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He dr...

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