Your roommate is working on his bicycle and has the bike upside down. He spins the 56.0 cm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second.
What is the pebble's speed?
What is the pebble's acceleration?
how far is one turn?
The circumference pi D = pi * 0.56 meter
so it goes 3 pi D /second
3 pi D m/s = 5.28 meters/second
Ac = V^2/R = 5.28^2 / (0.28) m/s^2
To find the pebble's speed, we need to determine how far it travels in one second. We know that the pebble goes by three times per second, so we need to calculate the distance it travels in one complete revolution.
1. First, find the circumference of the wheel using the formula: circumference = π * diameter.
Given:
Diameter = 56.0 cm
Circumference = π * 56.0 cm
Circumference ≈ 176.0 cm (rounded to one decimal place)
2. Now we know that the pebble travels the wheel's circumference (176.0 cm) three times in one second. To find the distance traveled in one second, multiply the circumference by the number of revolutions.
Distance traveled in one second = 3 * 176.0 cm
Distance traveled in one second = 528.0 cm (rounded to one decimal place)
The pebble's speed is 528.0 cm per second.
Next, to find the pebble's acceleration, we need to consider that the pebble is moving in a circular path and changing direction as the wheel spins. Therefore, we can use the centripetal acceleration formula:
Centripetal acceleration (a) = (v^2) / r
Given:
Speed (v) = 528.0 cm/s (from the previous calculation)
Radius (r) = diameter / 2 = 56.0 cm / 2 = 28.0 cm
Plugging in the values:
Centripetal acceleration (a) = (528.0 cm/s)^2 / 28.0 cm
Centripetal acceleration (a) ≈ 9,900.0 cm/s^2 (rounded to one decimal place)
Therefore, the pebble's acceleration is approximately 9,900.0 cm/s^2 (rounded to one decimal place).