Question

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its length is half the clock diameter. A light directly above the clock casts a shadow of the minute hand on the floor. The initial time on the clock is 15 15 minutes after noon. Let the positive x - x- axis points from the center of the clock towards the three o'clock position, and the positive y - y- axis point from the center of the clock towards the twelve o'clock position. Construct the algebraic expression for the length x ( t ) x(t) of the shadow of the minute hand as a function of time t .

Answer 1

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

• d is the length of the minute hand (d=D/2)
• D is the diameter of the clock
• t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!