You drive down the road at 39 m/s (88 mi/h) in a car whose tires have a radius of 34 cm.
What is the period of rotation of the tires?2sf value/units
Through what angle does a tire rotate in one second?2sf value/ units
period? Time=distance/2PIr=39m/s /.34*PI=
2PIr/39m/s= .0548sec
each rotation of 2PI occurs in that time, so in one second it
angle= 2PI*1/.0548 radians
i couldn't find the answer for part one still is wrong pls help
You drive down the road at 39 m/s (88 mi/h) in a car whose tires have a radius of 34 cm.
What is the period of rotation of the tires?2sf value/units
Through what angle does a tire rotate in one second?2sf value/ units
physics help - Steve, Saturday, April 29, 2017 at 4:46pm
each revolution travels 2πr = 68π cm. So,
3900cm/s * 1rev/68πcm = 3900/68π rev/s
That gives you frequency. Flip it upside down to get the period.
the angular speed is the linear speed divided by 2π.
physics help - bobpursley, Saturday, April 29, 2017 at 4:46pm
period? Time=distance/2PIr=39m/s /.34*PI=
2PIr/39m/s= .0548sec
each rotation of 2PI occurs in that time, so in one second it
angle= 2PI*1/.0548 radians
@bobpursley gets it right - Steve, Saturday, April 29, 2017 at 6:00pm
dang - I meant the angular speed is the rotational speed (rev/s) times 2π.
Well, well, well, if it isn't a tire-related question! Let's get rolling with the answers, shall we?
The period of rotation of the tires can be found by using the formula T = 2πr/v, where T is the period, r is the radius, and v is the velocity. Plugging in the given values, we have T = 2π(0.34 m)/(39 m/s). After crunching the numbers, the period of rotation of the tires is approximately 0.017 seconds (2 significant figures).
Now, let's find out through what angle does a tire rotate in one second. We can calculate this by using the formula θ = 2π/T, where θ is the angle and T is the period. Using the result from the previous calculation for T, we have θ = 2π/0.017 ≈ 369.7 radians (2 significant figures).
So, in one second, a car tire rotates through approximately 369.7 radians. Who knew tires were such spinning superstars?
To find the period of rotation of the tires, we can use the formula:
Period (T) = 2π * (radius of the tire) / linear velocity.
1. First, convert the radius of the tire from centimeters to meters:
Radius = 34 cm = 0.34 m
2. Convert the linear velocity from miles per hour (mi/h) to meters per second (m/s):
Velocity = 88 mi/h = (88 * 1609.34 m) / (60 * 60 s) ≈ 39.2 m/s
3. Substitute the values into the formula to calculate the period:
T = 2π * (0.34 m) / (39.2 m/s) ≈ 0.054 seconds (rounded to 2 significant figures)
Therefore, the period of rotation of the tires is approximately 0.054 seconds.
To find the angle through which a tire rotates in one second, we can use the formula:
Angle = (angular velocity) * (time).
The angular velocity can be determined as the ratio of linear velocity to the tire's radius:
Angular velocity = linear velocity / radius.
1. Divide the linear velocity by the radius to find the angular velocity:
Angular velocity = 39.2 m/s / 0.34 m ≈ 115.29 rad/s
2. Multiply the angular velocity by the time (1 second) to find the angle:
Angle = 115.29 rad/s * 1 s ≈ 115.3 radians (rounded to 2 significant figures)
Therefore, the tire rotates through approximately 115.3 radians in one second.
dang - I meant the angular speed is the rotational speed (rev/s) times 2π.
each revolution travels 2πr = 68π cm. So,
3900cm/s * 1rev/68πcm = 3900/68π rev/s
That gives you frequency. Flip it upside down to get the period.
the angular speed is the linear speed divided by 2π.