A long string is wrapped around a 6.4-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of 2.0m/s2 until 1.5m of string has been unwound. If the string unwinds without slipping, what is the cylinder's angular speed, in rpm, at this time?
To find the cylinder's angular speed, we need to consider the relationship between the linear speed of the string and the angular speed of the cylinder.
First, we find the linear speed of the string. Since the string is unwound without slipping, the linear speed of the string is equal to the linear acceleration times the distance of string unwound:
linear speed (v) = linear acceleration (a) * distance (d)
Given that the linear acceleration (a) is 2.0 m/s^2 and the distance (d) is 1.5 m, we can calculate the linear speed (v):
v = 2.0 m/s^2 * 1.5 m = 3.0 m/s
Next, we relate the linear speed of the string to the angular speed of the cylinder. The linear speed is equal to the product of the angular speed (ω) and the radius (r) of the cylinder:
linear speed (v) = angular speed (ω) * radius (r)
The radius (r) of the cylinder is half its diameter, so r = 6.4 cm / 2 = 3.2 cm = 0.032 m.
Plugging in the linear speed (v) and the radius (r), we can solve for the angular speed (ω):
3.0 m/s = ω * 0.032 m
ω = 3.0 m/s / 0.032 m = 93.75 rad/s
Finally, we convert the angular speed from radians per second to revolutions per minute (rpm). Since 1 revolution is equal to 2π radians, we can use the conversion factor:
ω (rpm) = ω (rad/s) * (60 s / 2π rad)
Plugging in the value for ω in rad/s, we can calculate the angular speed in rpm:
ω (rpm) = 93.75 rad/s * (60 s / 2π rad) = 890.25 rpm
Therefore, the cylinder's angular speed at this time is approximately 890.25 rpm.