A yo-yo with an axle diameter of 2.00cm has a 80.0cm length of string wrapped around it many times in such a way that the string completely covers the surface of its axle, but there are no double layers of string. The outermost portion of the yo-yo is 4.00 {\rm cm} from the center of the axle.If the yo-yo is dropped with the string fully wound, through what angle does it rotate by the time it reaches the bottom of its fall?
The circumference of the axel is
L=πd,
the string wraps around the axel
N=s/L= s/ πd.
The angle of rotation is
φ=2πN=2 πs/ πd=2s/d= =2•80/2=80 rad
To determine the angle of rotation, we need to consider the length of the string wrapped around the yo-yo's axle and the distance of the outermost portion of the yo-yo from the center of the axle.
The circumference of the yo-yo's axle is given by the formula:
C = 2πr
where 2π is approximately 6.28 and r is the radius of the axle. In this case, the radius is half of the axle diameter, so r = 2.00 cm / 2 = 1.00 cm.
Thus, the length of the string wrapped around the axle is 80.0 cm.
Next, we need to calculate the radius of the outermost portion of the yo-yo. The distance from the center of the axle to the outermost portion is given as 4.00 cm.
To find the angle of rotation, we can use the formula:
θ = arc length / radius
where θ is the angle of rotation, arc length is the length of the string wrapped around the axle, and radius is the radius of the outermost portion of the yo-yo.
Plugging in the values, we have:
θ = 80.0 cm / 4.00 cm
Simplifying further,
θ = 20
Therefore, the yo-yo rotates through 20 radians by the time it reaches the bottom of its fall.