How Do You Calculate Rotation Angles and Motion in Physics Problems?

[winterma]

Hi, I have a few questions, thanks for any help you can give!

1. If a wheel is rotating at a rate of 3.1 revolutions every 3.6 seconds. Through what angle, in radians, does the wheel rotate in 1 second??

2. A car can negotiate an unbanked curve safely at a certain maximum speed when the coefficient of static friction between the tires and the ground is 0.85. At what angle should the same curve be banked for the car to negotiate the curve safely at the same maximum speed without relying on friction?

3. A 0.073 kg arrow is fired horizontally. The bowstring exerts an average force of 60N on the arrow over a distance of 0.78m. With what speed does the arrow leave the bow? (for this one, i used the formula 1/2mv^2, but did not get the correct answer, maybe I'm using the incorrect formula)

Thanks!

 

[AKG]

You can figure these out yourself with a little help.

There is an equation very similar to the equation that relates constant speed, time, and distance traveled to an equation that relates similar concepts, but their rotational analogues. You know the equation:

$$v = \frac{\Delta d}{\Delta t}$$

where $v$ is the constant speed, $\Delta d$ is the distance travelled, and $\Delta t$ is the time elapsed. Now, let $\Delta \theta$ represent the angle through which an object has gone, and $\omega$ the constant angular velocity. There is a formula, then, as follows:

$$\omega = \frac{\Delta \theta}{\Delta t}$$

2. A car can negotiate an unbanked curve safely at a certain maximum speed when the coefficient of static friction between the tires and the ground is 0.85. At what angle should the same curve be banked for the car to negotiate the curve safely at the same maximum speed without relying on friction?

Coefficient of static friction? I don't exactly understand what it means to bank a curve at a certain angle.

3. A 0.073 kg arrow is fired horizontally. The bowstring exerts an average force of 60N on the arrow over a distance of 0.78m. With what speed does the arrow leave the bow? (for this one, i used the formula 1/2mv^2, but did not get the correct answer, maybe I'm using the incorrect formula)

Why did you use that formula? Do you know what that formula is for? Can you show your work? What is the book's given answer?

 

[M98Ranger]

1. To find the angle in radians, we can use the formula θ = (2π * n)/t, where θ is the angle, n is the number of revolutions, and t is the time in seconds. In this case, n = 3.1 revolutions and t = 3.6 seconds. Plugging in the values, we get θ = (2π * 3.1)/3.6 = 5.44 radians. Therefore, the wheel rotates by 5.44 radians in 1 second.

2. To find the angle at which the curve should be banked, we can use the formula tanθ = v^2/rg, where θ is the angle, v is the speed, r is the radius of the curve, and g is the acceleration due to gravity. In this case, we are given the coefficient of static friction, which is equal to tanθ. Therefore, we can substitute 0.85 for tanθ and solve for θ. We get θ = tan^-1(0.85) = 41.8 degrees. Therefore, the curve should be banked at an angle of 41.8 degrees.

3. To find the speed of the arrow, we can use the formula v = √(2Fd/m), where v is the speed, F is the force, d is the distance, and m is the mass of the arrow. Plugging in the values, we get v = √(2 * 60 * 0.78)/0.073 = 34.7 m/s. Therefore, the arrow leaves the bow with a speed of 34.7 m/s. It is important to note that the formula you used, 1/2mv^2, is the formula for kinetic energy and does not take into account the work done by the force of the bowstring. Using the correct formula will give you the correct answer.