What is the Angle Turned by the Wheel of a Car During Acceleration?

[chase23]

Alright, well, I tried and tried to solve this problem, but seemed to be getting no where.

A car travels at 1.70 m/s. The driver accelerates and increases the speed to 10.7 m/s in 2.20 s. If the radius of its wheel is 0.590 m, calculate the angle turned by the wheel during this time.

 

[Ouabache]

Welcome chase23 to physics forums!
There are a lot of helpful people here who would be happy to assist you, but you need to show your thoughts and where you are getting stuck, first.

Be sure to read this https://www.physicsforums.com/showthread.php?t=94379 too.

 

[chase23]

I know that I'll be using the formula angular displacement=time x angular velocity. The problem gave me the time, 2.20 s so now i need to find the angular velocity.

I believe the equation linear velocity = radius x angular velocity should give me angular velocity. THe problem also gave me radius: 0.590m. What I'm confused about is the "The driver accelerates and increases the speed to 10.7 m/s in 2.20 s. Is 10.7 the linear velocity? So 10.7 = 0.590 x angular velocity?

Anyways, I think I'm on the right track and need just a little guidance.

 

[mukundpa]

Whether the wheels are slipping or just rolling.

If only rolling there must be some relation between the distance covered and the angle turned.

Radius of the wheels is given.

 

[HallsofIvy]

The angular velocity is not constant- it depends on how fast the car is going. Assuming the wheels are rolling, rather than slipping, as mukundpa mentioned, which seems appropriate, remember that a tire tread covers exactly the same distance as the car. How far will the car go in 2.2 seconds? What is the circumference of a tire?

 

[maverick6664]

Assume acceleration is constant. This is not mentioned, but it's natural. And at first put aside angular velocity because you don't need it. All you need is how much angle the wheel proceeded.

Then the speed is given

$$v = v_0 + at$$ where a is the acceleration.

Now we get a.

$$v(2.2) = v_0 + 2.2a = 10.7m/s,$$ so $$a= \frac {10.7-1.7}{2.2} = 9/2.2$$

The distance the car advanced is given as

$$\int_0^{t_1} (v_0 + at)dt = v_0 t_1 + \frac 1 2 a t_1^2$$

and the angle the wheel rorated in radian is given as

$$\frac {v_0 t_1 * \frac 1 2 a t_1^2} {r} = 1304/59?$$

(strange value; my mistake? Anyway it's in radian. Formulae are correct )

 

[Astronuc]

A car travels at 1.70 m/s. The driver accelerates and increases the speed to 10.7 m/s in 2.20 s. If the radius of its wheel is 0.590 m, calculate the angle turned by the wheel during this time.

Determine the distance traveled during 2.2 s.

Now, the circumference (the distance traveled with one revolution, assuming no slip) of the circle is 2$\pi$r, where r is the radius, and 2$\pi$ is the angle (in radians) in one revolution.

Also, the '1304' should be '1364' according to my calcs.

 

[andrewchang]

this is just angular velocity...

$$v= \omega r$$

 

[chase23]

Alright, i was able to follow what you said maverick and got 1364/59 radians. That should be the answer, correct? I'll double check with my professor on tuesday, but would like to get a headstart on the problem.

 

[Gamma]

= 23.1 rad is what i get too.