Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

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Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

ss M-100 kg. In operation, the wheel rotates at 60.0 rev/min. While the wheel is spinning, your grandmother works clay at the center of the wheel with her hands into a pot-shaped object with circular symmetry. When the correct shape is reached, she wants to stop the wheel in as short a time interval as possible, so that the shape of the pot is not further distorted by the rotation. She pushes continuously with a wet rag as hard as she can radially inward on the edge of the wheel and the wheel stops in 6.00 s
(a) You would like to build a brake to stop the wheel in a shorter time interval, but you must determine the coefficient of friction between the rag and the wheel in order to design a better system. You determine that the maximum pressing force your grandmother can sustain for 6.00 s is 50.0N. k0.544

(b) What If? If your grandmother instead chooses to press down on the upper surface of the wheel a distance r 0.250 m from the axis of rotation, what is the force (in N) needed to stop the wheel in 6.00 s? Assume that the coefficient of kinetic friction between the wet rag and the wheel remains the same as before (Enter the magnitude.) 25.99 Remember that a torque is a product of a force and a distance. N

Physics

1 answer:

serg [3.4K] · Answer 1 · 2026-04-07T14:01:12+03:00

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is provided by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is determined by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction stands at 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

To halt the wheel, a force of 104.00902 N is required