c) A penny sits 12 cm from the center of a record turning at 33 1/3 rpm. What is the tangential speed in m/s?
d) A mass spins with a tangential speed of 1.2 m/s on the end of a string which is 0.75m long. What is the period of the rotation in seconds?
c. C = pi*2r = 3.14*24cm = 75.4cm=0.754 m.
V=33.33rev/min*0.754m/rev*1min/60s = 0.42 m/s.
To find the tangential speed in m/s for a penny sitting 12 cm from the center of a record turning at 33 1/3 rpm, we can use the following formula:
Tangential speed = 2πr × rotation speed
Where:
- π (pi) is a mathematical constant approximately equal to 3.14159
- r is the distance from the center (in meters)
- rotation speed is the number of revolutions per minute (rpm)
First, we need to convert the distance from centimeters to meters. Since 1 meter is equal to 100 centimeters, we divide 12 cm by 100 to get 0.12 meters.
Next, we need to convert the rotation speed from rpm to radians per second. Since there are 60 seconds in a minute, we divide 33 1/3 rpm by 60 to get the rotation speed in revolutions per second. Then, we multiply by 2π to convert from revolutions to radians.
Now, we can substitute the values into the formula:
Tangential speed = 2π × 0.12 × (33 1/3 ÷ 60)
Calculating this, we get:
Tangential speed ≈ 0.12 × (2π × 33.333 ÷ 60)
Tangential speed ≈ 0.12 × (2π × 0.55555)
Tangential speed ≈ 0.12 × (3.49158)
Tangential speed ≈ 0.4197916 m/s
Therefore, the tangential speed of the penny is approximately 0.4198 m/s.
Now let's move on to calculating the period of rotation for a mass spinning with a tangential speed of 1.2 m/s on the end of a string which is 0.75m long.
The period of rotation is the time it takes for one complete revolution. To find this value, we can use the following formula:
Period = circumference / tangential speed
The circumference is given by 2πr, where r is the length of the string.
Substituting the values into the formula, we have:
Period = (2π × 0.75) / 1.2
Simplifying this expression, we get:
Period ≈ 4.71238898 / 1.2
Calculating this, we find:
Period ≈ 3.92699082
Therefore, the period of rotation is approximately 3.927 seconds.