A hockey puck of diameter 3 inches is spinning around its center at a speed of 3 counterclockwise rotations per second. The center of the puck is traveling at a speed of 24 inches per second at an angle of 45 degrees to the positive real axis.
(a) At time t=0, the center of the puck is at the origin. Where is the center of the puck at time t? (Time t is measured in seconds.)
(b) A point on the outer edge of the puck begins at the point (0, 3/2). What is its location at time t?
I tried solving this problem but I don't know where to start. This almost looks like a physics problem. How do you incorporate the "spinning"?
(a) the horizontal and vertical speeds are 24/√2 = 6√2, so at time t, the center is at
(6√2 t, 6√2 t)
(b) the puck is a circle of radius 3/2, and angular velocity = 3*2π /s
The point (0,3/2) corresponds to θ=0
So, we have, relative to the puck center,
x = 3/2 cosθ
y = 3/2 sinθ
But θ = 6πt, so point P on the puck has coordinates
x = 3/2 cos(6πt) + 6√2 t
y = 3/2 sin(6πt) + 6√2 t
Thanks, but I was wondering how you got the horizontal and vertical speeds, sorry I'm not that good at parametric equations yet :( Also how did you get that the angular velocity is 3*2pi / s?
Thanks
2pi radians per rotation. 3 rotations/sec
Oops. (0,3/2) corresponds to θ=π/2, so
x = 3/2 cos(6πt+π/2) + 6√2 t
y = 3/2 sin(6πt+π/2) + 6√2 t
or
x = -3/2 sin(6πt) + 6√2 t
y = 3/2 cos(6πt) + 6√2 t
Well, I would just like to point out to steve that 24/√2 = 12√2, not 6√2.
And Samuel? Ahem.
Yes, this problem does involve some physics concepts. To tackle it, we need to consider the linear and rotational motion of the hockey puck.
To start, let's analyze the spinning motion of the puck. The puck completes 3 counterclockwise rotations per second. This means it completes one full revolution (360 degrees) in \(\frac{1}{3}\) of a second.
The circumference of the puck can be calculated using its diameter, which is given as 3 inches. The formula to find the circumference of a circle is \(C = \pi \times d\), where \(d\) is the diameter. Substituting in the given values, we get \(C = \pi \times 3\). Therefore, the circumference of the puck is \(3\pi\) inches.
Now, let's consider the linear motion of the center of the puck. It is traveling at a speed of 24 inches per second at an angle of 45 degrees to the positive real axis. We can break down this motion into horizontal and vertical components.
The horizontal component of the linear motion can be found using cosine of the angle. So, the horizontal velocity (Vx) is \(24 \times \cos(45^\circ)\).
The vertical component of the linear motion can be found using sine of the angle. So, the vertical velocity (Vy) is \(24 \times \sin(45^\circ)\).
(a) At time \(t = 0\), the center of the puck is at the origin. To determine its position at time \(t\), we need to consider both the linear and rotational motion. The center of the puck moves linearly with the velocity components Vx and Vy, and it also completes multiple rotations.
To incorporate the spinning, we can calculate the angular displacement, which is the amount of rotation in radians. Since the puck completes one full revolution in \(\frac{1}{3}\) of a second, the angular displacement is given by \(\theta = 2\pi \times \frac{t}{\frac{1}{3}}\). Simplifying this expression, we have \(\theta = 6\pi t\).
Using the angular displacement, we can find the position of the center of the puck as follows:
- The x-coordinate of the center is given by \(x = Vx \times t\).
- The y-coordinate of the center is given by \(y = Vy \times t\).
Therefore, the position of the center of the puck at time \(t\) is \((Vx \times t, Vy \times t)\).
(b) For a point on the outer edge of the puck, we need to account for both the linear and rotational motion. The outer edge of the puck moves along the circumference of the puck while also following linear motion.
Let's consider the point (0, 3/2) at time \(t = 0\). As the puck rotates, this point will have traveled a certain distance along the circumference of the puck, which can be derived from the angular displacement.
Using the angular displacement \(\theta = 6\pi t\), we can calculate the arc length traveled by the point (0, 3/2) as follows:
- Arc length = \(\theta \times \text{radius}\), where the radius is half of the diameter of the puck.
- Arc length = \(6\pi t \times \frac{3}{2}\).
Now, we have the distance traveled along the circumference. However, we also need to consider the linear motion of the center. The distance traveled linearly by the point (0, 3/2) can be calculated by multiplying the linear velocity by time, which is given by \(24 \times t\).
To obtain the final position of the point, we need to combine the linear and rotational motion. The x-coordinate of the point is given by:
- x-coordinate = \(Vx \times t + \text{arc length}\).
The y-coordinate of the point is given by:
- y-coordinate = \(Vy \times t\).
Therefore, the location of the point on the outer edge of the puck at time \(t\) is \((Vx \times t + \text{arc length}, Vy \times t)\).