Another Rotational Motion Question

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SUMMARY

The discussion focuses on calculating the force exerted by a tradesman sharpening a knife against a grindstone. The grindstone has a diameter of 35 cm, spins at 200 rpm, and has a mass of 15 kg, with a coefficient of kinetic friction of 0.20. After determining the angular velocity and the normal force, the user calculates the frictional force to be 29.4 N. The conversation emphasizes the importance of understanding torque, moment of inertia, and angular acceleration in solving the problem.

PREREQUISITES
  • Understanding of rotational motion concepts, including torque and angular acceleration.
  • Familiarity with the equations of motion, specifically Sigma F = ma and F = μ_k * N.
  • Knowledge of converting rotational speed from rpm to rad/s.
  • Basic grasp of frictional forces and their calculations in physics.
NEXT STEPS
  • Study the relationship between torque and moment of inertia in rotational dynamics.
  • Learn how to calculate angular acceleration and its impact on rotational motion.
  • Explore the effects of friction in different materials and their coefficients.
  • Investigate practical applications of rotational motion in mechanical systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, tradespeople involved in sharpening tools, and anyone interested in the principles of rotational motion and friction.

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Homework Statement


A tradesman sharpens a knife by pushing it against the rim of a grindstone. The 35 cm diameter stone is spinning at 200 rpm and has a mass of 15 kg. The coefficient of kinetic friction between the knife and the stone is 0.20.

If the stone loses 10% of its speed in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Homework Equations


SIgma F=ma
F=mu_k*N
A_c= (Mv^2/r)
I would assume that because the knife is pushed into the stone force should be counted as centripetal force.

The question asked for N for units and values

The Attempt at a Solution



200 rpm * 2pi/60s = 20.944 rad/s
V = 20.944*.175 = 3.6652 m/s
N= 9.8*15= 147N

Velocity With Speed Decay = 3.2987 m/s
F=147*.2= 29.4N

I can't think of anywhere to go from here, and any sort of help would be appreciated.
 
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Try it top-to-bottom. The big picture is
Torque = moment of inertia x angular acceleration

If you fill in the details on each of those parts and play around with it a bit you should be able to solve for the applied force.
 

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