A grindstone of radius 10 cm and moment of inertia 0.2 kg: m? rotates at 200 rpm. A tool is pressed against the rim with force of 50 N along radial direction. The coefficient of kinetic friction is 0.6. (a) What is the power needed to keep the stone rotating at fixed rate? (b) If the drive is removed, how long How would it take for the stone to stop? (c) many revolutions does it make while slowing down?
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(a) To calculate the power needed to keep the grindstone rotating at a fixed rate, we can use the formula for rotational power:
\[P = \tau\omega\]
Where:
P = power
τ (tau) = torque
ω (omega) = angular velocity
First, calculate the torque. The torque due to the force applied is given by:
\[τ = rF\]
Where:
r = radius of the grindstone = 10 cm = 0.1 m
F = force applied along the radial direction = 50 N
\[τ = 0.1 m × 50 N = 5 N·m\]
Next, we need to convert the angular velocity from rpm to rad/s:
\[ω = \frac{2π × 200}{60}\]
Now, we can calculate the power:
\[P = 5 N·m × \frac{2π × 200}{60} \text{ rad/s} = 1047.2 \text{ W} (watts)\]
So, the power needed to keep the stone rotating at a fixed rate is approximately 1047.2 watts.
(b) To find out how long it would take for the stone to stop when the drive is removed, we can use the concept of rotational kinetic energy. The work-energy theorem states that the change in kinetic energy is equal to the work done on the object.
The initial kinetic energy of the grindstone is:
\[KE_{\text{initial}} = \frac{1}{2}Iω^2\]
Where:
I = moment of inertia = 0.2 kg·m²
ω = angular velocity in rad/s (we already calculated it)
The final kinetic energy is zero (as it comes to a stop).
The work done on the stone is equal to the change in kinetic energy:
\[W = KE_{\text{final}} - KE_{\text{initial}}\]
Since the final kinetic energy is zero, we have:
\[W = -\frac{1}{2}Iω^2\]
The negative sign indicates work is done against the stone's motion.
Now, we can calculate the time it takes for the stone to stop using the work-energy theorem:
\[W = Fd\]
Where:
F = frictional force = μN
N = normal force
d = distance over which work is done
Since the force is applied along the radial direction, N = mg (where m is the mass of the stone, and g is the acceleration due to gravity).
Now, we need to calculate the distance d over which the frictional force does work. The grindstone will come to a stop, so we'll be looking for the distance when its final kinetic energy becomes zero.
Using the work-energy theorem:
\[W = Fd = -\frac{1}{2}Iω^2\]
We can solve for d:
\[d = -\frac{Iω^2}{2F}\]
Now, we can calculate the time it takes to stop:
\[W = Fd\]
\[d = \frac{W}{F}\]
And we already have the expression for d in terms of I, ω, and F.
Now we can calculate the time it takes for the stone to stop:
\[t = \frac{d}{v}\]
Where:
d = distance
v = final velocity (0, since it stops)
So, find the time t.
(c) To calculate the number of revolutions while slowing down, we can use the relationship between linear distance and angular distance. The linear distance traveled during the stopping process is equal to the circumference of the grindstone (2πr) times the number of revolutions.
So, you can find the linear distance d using:
\[d = 2πr \times \text{number of revolutions}\]
We already calculated the distance d when calculating the time it takes to stop. So, use that value to find the number of revolutions.