Answer:
2.09.ft/min
Step-by-step explanation:
There is a formula for the linear velocity using rpm:
V = (2π)/60 * r * rpm
V = (2π)/60 * 2ft * 10 = 2.09ft/min
Answer:
Linear velocity of point on the edge of wheel having radius of 2 feet and spin rate of 10 revolution per minute is 125.71 feet per minute.
Solution:
Given that radius of wheel = 2 feet
There is a point on edge of the wheel .we need to determine linear velocity of that point.
Let’s first calculate distance covered by a point when 1 revolution of wheel is complete.
When one revolution is complete the distance traveled by a point on edge of the wheel will be equal to circumference of the wheel
[tex]=2 \pi \mathrm{r}=2 \mathrm{x}\left(\frac{22}{7}\right) \times 2=\frac{88}{7} \mathrm{feet}[/tex]
In one revolution, point covers distance of [tex]\frac{88}{7}[/tex] feet
So in 10 revolution, point covers distance of [tex]\frac{88}{7} \times 10 = \frac{880}{7}[/tex]
Given that in a minute, wheel takes 10 revolution.
Which means in a minute , point covers [tex]\frac{880}{7}[/tex] feet that is [tex]\frac{880}{7}[/tex] feet per minute = 125.71 feet per minute
Hence linear velocity of point on the edge of wheel having radius of 2 feet and spin rate of 10 revolution per minute is 125.71 feet per minute.
