Relating linear and angular kinematics

In summary, a wheel with a radius of 0.200m has a tangential speed of 50.0 m/s at t=3.00s and is slowing down with a constant tangential acceleration of 10.0 m/s^2. The constant angular acceleration is -50 radian/s^2 and the constant angular velocities at t=3.00s and t=0 are 250 radians/s and 400 radians/s, respectively. The wheel turns through an angle of 975 radians between t=0 and t=3.00s. The wheel will have a radial acceleration equal to g when t=0.12s. To solve this problem, you can use the formula R omega^2=
  • #1
kenren
im really having a hard tym doing this assignment of mine, i hope someone cud help me out on this one.

at t= 3.00s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the whell slows down with a tangential acceleration of constant magnitude 10.0 m/s2.
a.)Calculate the wheel's constant angular acceleration.
b.) calculate the wheel's constant angular velocities at t= 3.00s and t= 0.
c.) through what angle did the wheel turn between t=0 and t=3.00s?
At what time will the radial acceleration equal g?


can you tell me the formula and the process on how to solve this problems..please.. I've tried to solve this one but just couldn't get any good answers. please neone help me...:frown:
 
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  • #2
An entire circle is 2pi radians and the circumference of a circle of radius 0.2 m is .4 pi m. That means that an angular velocity of omega radians/s is the same as (1/(2pi)omega) revolutions per second or (.4pi)(1/(2pi))omega = .2 omega m/s tangential speed. Since you are given that, at t=3, the tangential speed is 50 m/s, it follow that the angular velocity at that time satisfies 50= .2 omega or
omega= 50/.2= 250 radians/s. (That's part of the answer to part b.)

If the wheel was slowing down at a constant 10 m/s^2, then, when t=0, the tangential speed must have been 50+ 3(10)= 80 m/s^2. By the same argument, the angular velocity at that time was 80/.2=
400 radian/s. (That's the rest of the answer to part b.)

Since the angular velocity changed from 400 radians/s to 250 radians/s in 3 seconds, the angular acceleration is (250-400)/3=
-150/3= -50 radians/s^2. (That's the answer to part a.)

For a constant angular acceleration of -50 m/s and angular velocity at t=0 of 400 radians/sec, the angular velocity after t seconds is 400- 50t radians/sec. The "position" (in the sense of angles) is the integral of that 400t- 25t^2 (assuming 0 as the initial "angle") so in 3 seconds, the wheel will have turned through an angle of 400(3)- 25(3^2)= 1200- 25(9)= 1200- 225= 975 radians. (That's the answer to part c.)

If you don't want to use calculus (the "integral") you can use an "averaging method". At t=0, the angular velocity is 400 radians/s and at t= 3, the angular velocity is 250 radians/sec. The average angular velocity is (400+ 250)/2= 650/2= 325 radians/sec (averaging using only the first and last values works ONLY when the rate of change (here acceleration) is constant). Moving at 325 radians/s for 3 sec means the wheel moves through 325*3= 975 radians as above. (Of course, proving that, for constant acceleration, we can average like that, requires calculus!)

The radial acceleration of a wheel with radius R and angular velocity of omega is R omega^2. That will be equal to g when, of course, R omega^2= g. Here, R= 0.2, g= 9.8, and omega= 400- 50t so you need to solve 0.2(400-50t)^2= 9.8 to answer the last question.
 
  • #3


Linear and angular kinematics are two ways of describing the motion of an object. Linear kinematics deals with the motion of an object in a straight line, while angular kinematics deals with the motion of an object in a circular path. These two types of motion are related through the concept of radius and angular displacement.

To relate linear and angular kinematics, we can use the formula v = rω, where v is the linear velocity, r is the radius, and ω is the angular velocity. This formula shows that the linear velocity of a point on the rim of a wheel is directly proportional to its distance from the center of the wheel (radius) and its angular velocity.

Now, let's apply this concept to the given problem. At t=3.00s, the point on the rim of the wheel has a tangential speed of 50.0 m/s and a radius of 0.200m. Using the formula v = rω, we can solve for the angular velocity at t=3.00s:

50.0 m/s = (0.200m)ω

ω = 50.0 m/s / (0.200m)

ω = 250 rad/s

a.) To calculate the wheel's constant angular acceleration, we can use the formula ω = ω0 + αt, where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time. Since we know the initial angular velocity at t=0 is 0 rad/s, we can solve for α:

250 rad/s = 0 rad/s + α(3.00s)

α = 250 rad/s / 3.00s

α = 83.3 rad/s^2

b.) To calculate the wheel's constant angular velocity at t=3.00s, we already solved for it in part a. At t=0, the angular velocity is 0 rad/s.

c.) To find the angle through which the wheel turned between t=0 and t=3.00s, we can use the formula θ = θ0 + ω0t + 1/2αt^2, where θ0 is the initial angular displacement, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time. Since θ0 = 0 and ω0 = 0, we can solve for θ
 

1. How are linear and angular kinematics related?

The relationship between linear and angular kinematics can be described by the concept of rotational motion. Rotational motion is the movement of an object around an axis or point. The linear displacement of an object can be converted to angular displacement by dividing the distance traveled by the radius of the circle it is moving in.

2. What is the difference between linear and angular kinematics?

The main difference between linear and angular kinematics is the type of motion they describe. Linear kinematics deals with the movement of objects in a straight line, while angular kinematics deals with the movement of objects in a circular or rotational path. Additionally, linear kinematics uses distance, velocity, and acceleration to describe motion, while angular kinematics uses angle, angular velocity, and angular acceleration.

3. How is angular velocity related to linear velocity?

Angular velocity and linear velocity are related through the radius of the circle an object is moving in. Angular velocity is defined as the rate of change of angular displacement, while linear velocity is defined as the rate of change of linear displacement. The two are related by the equation v = ωr, where v is linear velocity, ω is angular velocity, and r is the radius of the circle.

4. Can linear and angular kinematics be applied to real-life situations?

Yes, linear and angular kinematics can be applied to real-life situations. For example, the motion of a Ferris wheel involves both linear and angular kinematics. The cars on the Ferris wheel travel in a circular path (angular motion), but also have a linear velocity as they move up and down the wheel. The principles of linear and angular kinematics can also be applied to sports, such as the motion of a basketball being thrown into a hoop.

5. How is torque related to linear and angular kinematics?

Torque is a measure of the force that causes an object to rotate about an axis. It is related to both linear and angular kinematics because it is affected by the distance from the axis of rotation (linear displacement) and the angle at which the force is applied (angular displacement). This means that torque can be used to calculate both linear and angular acceleration, which are important factors in describing the motion of an object.

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