A peg on a turntable moves with a constant linear speed of 0.42 m/s in a circle of radius 0.25 m. The peg casts a shadow on a wall. Find the following quantities related to the motion of the shadow.
(a) the period
s
(b) the amplitude
m
(c) the maximum speed
m/s
(d) the maximum magnitude of the acceleration
m/s2
please help me! i dont understand how to do any of these. thanks so much!!
(a) the period
s
(b) the amplitude
m
(c) the maximum speed
m/s
(d) the maximum magnitude of the acceleration
m/s2
please help me! i dont understand how to do any of these. thanks so much!!
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I believe that the shadow in this case is what the motion of the peg would look like if you were to turn the circle so that you see only its side. Imagine, then, a line along which the shadow will oscillate as the peg rotates around the circle.
a) T=2πR/v=2π(0.25)/0.42=3.74s. To see why T=2πR/v, rearrange the equation to Tv=2πR. Now, we know that, by definition, T is the time required to complete one full rotation. Therefore, Tv should be the distance traveled in one full rotation, which is just 2πR.
b) A=R=0.25m. The amplitude is the maximum distance the shadow travels from the midpoint of the line, which is just the radius of the circle.
c) vmax=v=0.42m/s. When the shadow is at the midpoint of the line, the linear velocity has either a zero x-component or zero y-component (depending upon how you decide to set up you coordinate system), and the velocity of the shadow will be equal to the linear velocity. At other points on the line, the velocity of the shadow will decrease because the linear velocity, which is constant, will gain a component that is not reflected in the motion of the shadow.
d) a=(v^2)/R for uniform circular motion. The acceleration of an object in uniform circular motion is constant and is always pointed toward the center of the circle. When the shadow is at the midpoint of the line, we can't observe any acceleration because it is pointed into the page. This makes sense because, the velocity of the shadow is max at the midpoint. On the other hand, when the shadow is at either end of the line, we observe the entire length of the acceleration vector. Therefore, we can say that amax=(v^2)/R=0.7056m/s^2.
a) T=2πR/v=2π(0.25)/0.42=3.74s. To see why T=2πR/v, rearrange the equation to Tv=2πR. Now, we know that, by definition, T is the time required to complete one full rotation. Therefore, Tv should be the distance traveled in one full rotation, which is just 2πR.
b) A=R=0.25m. The amplitude is the maximum distance the shadow travels from the midpoint of the line, which is just the radius of the circle.
c) vmax=v=0.42m/s. When the shadow is at the midpoint of the line, the linear velocity has either a zero x-component or zero y-component (depending upon how you decide to set up you coordinate system), and the velocity of the shadow will be equal to the linear velocity. At other points on the line, the velocity of the shadow will decrease because the linear velocity, which is constant, will gain a component that is not reflected in the motion of the shadow.
d) a=(v^2)/R for uniform circular motion. The acceleration of an object in uniform circular motion is constant and is always pointed toward the center of the circle. When the shadow is at the midpoint of the line, we can't observe any acceleration because it is pointed into the page. This makes sense because, the velocity of the shadow is max at the midpoint. On the other hand, when the shadow is at either end of the line, we observe the entire length of the acceleration vector. Therefore, we can say that amax=(v^2)/R=0.7056m/s^2.