Asked by anonymous

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
Answered by Steve
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.
Answered by bobpursley
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
Answered by Steve
dang - I meant the angular speed is the rotational speed (rev/s) times 2π.
Answered by anonymous
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.

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