Total acceleration at the bottom of a rolling wheel

In summary, the buggy experiences an average acceleration of 0.9375 m/s^2 due to the wheel slipping on the ice.
  • #1
Angelique
3
0
A tundra buggy, which is a bus fitted with oversized wheels, is stuck in Churchill, Manitoba, on slippery ice. The wheel radius is 0.84 m. The speedometer goes from 0 to 27 km/h while the buggy moves a total distance of 7.0 m in 9.0 s.

Find the magnitude of the total acceleration of a point at the bottom of the wheel at the end of 4.0 s.

I know that the total acceleration is the tangential acceleration + the radial acceleration + the linear acceleration. But i can't seem to get the right values for the accelerations since the answer is 13.2m/s^2

for tangential acceleration i got 0.66m/s^2
Im not sure how to find the other 2 accelerations
 
Physics news on Phys.org
  • #2
How do speedos work?
What does the speedo actually measure? (Hint: not the overland speed.)
 
  • #3
It measures the speed of the wheel (angular speed) so then the angular acceleration is: (final angular velocity)/time? so 7.5/8 = 0.9375m/s^2? but that seems very small to me because linear acceleration is radius times the angular acceleration. :/
 
  • #4
Angelique said:
linear acceleration is radius times the angular acceleration.
Not when the wheel is slipping on the ice.
 
  • #5
It measures the speed of the wheel (angular speed) so then the angular acceleration is: (final angular velocity)/time? so 7.5/8 = 0.9375m/s^2? but that seems very small to me because linear acceleration is radius times the angular acceleration. :/
The speedo reads the vehicle speed in the event there is no slipping.
So it is calibrated to tell you the tangential speed of the wheel rim.
So the outer edge of the wheel goes from 0 to v in time T, giving an average angular acceleration of ##\alpha_{ave} = v/rT##
... notice that ##a_T=r\alpha=v/T## as expected?
(It is best practice to do the algebra with symbols and substitute the numbers in later.)

Note: that's just an example - you may need to consider more than the average acceleration. i.e. is the acceleration the same throughout the motion? What part does the distance moved play?
I'm guessing you have some notes about rolling with slipping?
 
Last edited:

What is total acceleration at the bottom of a rolling wheel?

The total acceleration at the bottom of a rolling wheel is the combined effects of both translational and rotational acceleration. It takes into account the linear acceleration of the center of mass and the angular acceleration due to the rotation of the wheel.

How is total acceleration at the bottom of a rolling wheel calculated?

Total acceleration at the bottom of a rolling wheel can be calculated using the formula a = (r * alpha) + (acm), where r is the radius of the wheel, alpha is the angular acceleration, and acm is the linear acceleration of the center of mass.

What factors affect the total acceleration at the bottom of a rolling wheel?

The total acceleration at the bottom of a rolling wheel is affected by the radius of the wheel, the angular acceleration, and the linear acceleration of the center of mass. Other factors that can affect it include the mass and shape of the wheel, as well as any external forces acting on the wheel.

How does the total acceleration at the bottom of a rolling wheel affect the motion of the wheel?

The total acceleration at the bottom of a rolling wheel determines the rate of change of the wheel's velocity and direction of motion. It also affects the stability of the wheel and its ability to maintain a constant speed and direction of motion.

Why is it important to consider the total acceleration at the bottom of a rolling wheel?

Understanding and calculating the total acceleration at the bottom of a rolling wheel is important in many practical applications, such as designing vehicles and machinery. It also helps to explain the complex motion of rolling objects and can be used to analyze and improve their performance.

Similar threads

  • Mechanical Engineering
Replies
8
Views
4K
  • Introductory Physics Homework Help
2
Replies
65
Views
4K
Replies
7
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
2
Views
4K
Replies
7
Views
3K
  • Mechanics
Replies
1
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Back
Top