Rotational kinematics of circular saw

[MozAngeles]

Homework Statement

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4740 rpm to zero in 3.00 s.
A. What is the angular acceleration of the blade(rev/s^2)? GOT IT
B. What is the distance traveled by a point on the rim of the blade during the deceleration(ft)? GOT IT
C. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration(in)? NEED HELP

Homework Equations

$$\theta$$=s/r
all of the angular kinematics equations

The Attempt at a Solution

i solved part be with s=$$\theta$$*r after i had found theta using the kinematics equation ($$\omega$$^2=$$\omega$$0^2+2$$\alpha$$*$$\Delta$$$$\theta$$, solving for theta using the acceleration from part A. IF you think part C is just the answer to part B in inches that's wrong, but I do not know what else to do. PLease help.

 

[Doc Al]

Imagine a circle. The top of the circle represents the starting point. Where is the end point of its motion? Draw an arrow between those points. That's the displacement.

 

[MozAngeles]

so the net displacement is the diameter of the wheel, right? So then to i use the S- theta*r, but for r use 10?

 

[Doc Al]

so the net displacement is the diameter of the wheel, right?

Only if the net angular displacement is an odd multiple of pi radians. In other words, only if the point of the blade ends up on the opposite side of the circle after it stops.

So then to i use the S- theta*r, but for r use 10?

Not sure what you mean here.

Think about this: How many revolutions does the blade make? If it makes one complete revolution, the net displacement is zero. It's back where it started. If it makes 2.5 revolutions, it is 180 degrees away, thus the displacement equals the diameter. And so on.

Make sense?

 

[rwisz]

I would approach this problem by first defining the variables and equations needed to solve it. In this case, we are dealing with rotational kinematics, so we can use the following equations:

1. \omega = \omega_0 + \alpha t
2. \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2
3. \omega^2 = \omega_0^2 + 2 \alpha \theta

Where:
- \omega is the angular velocity in radians per second
- \omega_0 is the initial angular velocity in radians per second
- \alpha is the angular acceleration in radians per second squared
- t is the time in seconds
- \theta is the angular displacement in radians
- \theta_0 is the initial angular displacement in radians

Now, let's solve for part A. We are given the initial and final angular velocities, as well as the time, so we can use equation 1 to find the angular acceleration:

\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - (4740 \text{ rpm}) \times (2\pi \text{ rad/rpm})}{3.00 \text{ s}} = -318.3 \text{ rad/s}^2

For part B, we need to find the distance traveled by a point on the rim of the blade during the deceleration. Since we know the angular acceleration, we can use equation 2 to find the angular displacement:

\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + (4740 \text{ rpm}) \times (2\pi \text{ rad/rpm}) \times (3.00 \text{ s}) + \frac{1}{2} (-318.3 \text{ rad/s}^2) (3.00 \text{ s})^2 = -3581 \text{ rad}

Now, we can use the relationship \theta = s/r to find the distance traveled by a point on the rim of the blade:

s = \theta r = (-3581 \text{ rad}) \times (10.0 \text{ in}/2) = -17905 \text{ in}

Since distance cannot be negative, we