Rotational speed versus linear motion problem

In summary, the speed of the bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance of 0.921 meters apart. From the angular displacement of the two bullet holes in the disks and the rotational speed of the disks, it can be determined that the bullet is traveling at a speed of 1330.544749 meters per second.
  • #1
Enoch
20
0
Here is the question:

The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 0.921 m apart on the same axle. From the angular displacement 27.4 degrees of the two bullet holes in the disks and the rotational speed 1050 rev/min of the disks, we can determine the speed v of the bullet. What is the speed of the bullet in units of m/s?

Okay to start the problem, I wrote out my known values:

d = o.921 m
r = 0.4605 m
initial angular velocity = 1050 rev/min
initial angular velocity = 1050 (rev/min) (1 min/ 60 sec) (2 x 3.14 rad/rev)
initial angular velocity = 109.9557429 rad/s
Theta = 27.4 degrees X ( 1 rev/ 360 degrees) = 0.0761111111

Those are my known values and their derivations. To solve the problem, I figured the speed of the bullet was reliant upon where the hole would be in the second disk. Thus I set up the equation:

theta = (1/2) alpha (t^2) + (initial angular velocity X t)
- Here I assumed alpha was 0, since I assumed the bullet was moving at a constant speed. Therefore:

theta = (initial angular velocity X t)
t = (0.0761111111/109.9557429)
t = 0.000692197688 seconds

With the time value found for the second disk, I figured that the bullet had to trave the diameter at that time. In other words, velocity is equal to diameter/time.

V = d/t
V = 0.921 m / 0.0006921976888 s
V = 1330.544749 m/s

This was my final answer, yet it is incorrect. Can anyone point out what I have done wrong and how to fix it? Sorry for the inconvienince.
 
Physics news on Phys.org
  • #2
I have'nt checked your arithemtic, but I'd have marked it wrong just because of the excessive number of digits you used. Read up on significant digits.
 
  • #3
If you convert to radians do it everywhere, you did not express the angle between the holes in radians, you have simply expressed them as a fraction of the full 360deg circle. Multiply your angle(.07...) by 2 [itex] \pi [/itex] that should help your answer.

As I said before, save yourself time and effort, do not even write down non significant digits.
 
  • #4
Integral said:
I have'nt checked your arithemtic, but I'd have marked it wrong just because of the excessive number of digits you used. Read up on significant digits.

The answer is supposed to be in 6 significant digits, which I answered the problem with. I just posted the numbers in their entirety here, in case I needed to use them to find the answer eventually. There is something else wrong with the problem unfortunately.
 
  • #5
Integral said:
If you convert to radians do it everywhere, you did not express the angle between the holes in radians, you have simply expressed them as a fraction of the full 360deg circle. Multiply your angle(.07...) by 2 [itex] \pi [/itex] that should help your answer.

As I said before, save yourself time and effort, do not even write down non significant digits.

I'll do that and get back to you guys on whether or not it worked ;).
 
  • #6
Wow, all that work, and I just missed such a little thing. Thanks Integral ;). It oftentimes takes somone else to point out the careless mistakes. I really appreciate the help.
 
  • #7
I think you over complicated the problem a bit, doing the check I went straight to deg/s then used that, with the 27.4 deg, to get the time of flight.
 
  • #8
Integral said:
I think you over complicated the problem a bit, doing the check I went straight to deg/s then used that, with the 27.4 deg, to get the time of flight.

Ah...No need to convert to radians then? I just remember my prof. hammering into us to work with radians :D.

The changes this causes, for those who look at this problem later are as follows:

theta = 0.47822 radians
time = 0.0043492 seconds
Velocity = 211.763 m/s
 
Last edited:
  • #9
Converting to radians is fine, in fact a very good habit.
 
  • #10
I dpn't understand why you converted to radians, since the only angle is given in degrees.

Here's how I would have done it:

The papers are rotating at 1050 revolutions per minute which is 1050*360= 378000 degrees per minute and therefore 378000/60= 6300 degrees per second.

The total rotation in T seconds is 6300T degrees. Since we are told that the papers rotated 27.4 degrees in the time the bullet was moving between the two papers, we must have 6300T= 27.4 or T= 27.4/6300= 0.003921 seconds.

Taking v to be the speed of the bullet, v= .921m/0.003921 seconds= 234.9 m/s

I also don't understand why you say "The answer is supposed to be in 6 significant digits" when the distance is given to only 3 significant digits. I would give the answer as 235 meters (or, better, 2.35x102 m.).
 
  • #11
HallsofIvy said:
I dpn't understand why you converted to radians, since the only angle is given in degrees.

Here's how I would have done it:

The papers are rotating at 1050 revolutions per minute which is 1050*360= 378000 degrees per minute and therefore 378000/60= 6300 degrees per second.

The total rotation in T seconds is 6300T degrees. Since we are told that the papers rotated 27.4 degrees in the time the bullet was moving between the two papers, we must have 6300T= 27.4 or T= 27.4/6300= 0.003921 seconds.

Taking v to be the speed of the bullet, v= .921m/0.003921 seconds= 234.9 m/s

I also don't understand why you say "The answer is supposed to be in 6 significant digits" when the distance is given to only 3 significant digits. I would give the answer as 235 meters (or, better, 2.35x102 m.).

The place I turn in the homework to always wants the answers in 6 significant digits...not the correct amount ;).
 
  • #12
wow, that is horrible that any professor would have their students do that...such horrible practice...
 
  • #13
Spectre5 said:
wow, that is horrible that any professor would have their students do that...such horrible practice...

Its an online homework thing that our professor uses. For our classwork, we actually have to use correct significant digits, but the homework agency doesn't "believe" in significant digits. Therefore, they assigned an arbitrary number...6...for all answers to come in. It makes things easier for the homework, but you have to watch yourself in the exams!

:rofl:
 

1. What is the difference between rotational speed and linear motion?

Rotational speed refers to the rate at which an object rotates around its axis, while linear motion is the movement of an object in a straight line. Rotational speed is measured in rotations per minute or radians per second, while linear motion is typically measured in meters per second.

2. How are rotational speed and linear motion related?

Rotational speed and linear motion are related through the radius of rotation. The linear speed of an object at a given rotational speed is directly proportional to the radius of rotation. This means that the larger the radius, the higher the linear speed at the same rotational speed.

3. Which type of motion is more efficient for energy transfer?

Generally, linear motion is more efficient for energy transfer compared to rotational motion. This is because linear motion involves less energy loss due to friction and other factors. However, this can vary depending on the specific situation and the design of the system.

4. How does rotational speed affect the stability of an object?

Rotational speed can greatly affect the stability of an object. As the rotational speed increases, the object's stability decreases, making it more prone to tipping or falling. This is why objects with high rotational speeds, such as a spinning top, often have a lower center of mass to maintain stability.

5. What are some real-life examples of rotational speed versus linear motion?

There are many real-life examples of rotational speed versus linear motion, such as a spinning top or a ceiling fan. Another example is a car's engine, which converts linear motion of the pistons into rotational motion to power the wheels. In sports, a pitcher's throw is also a combination of rotational speed and linear motion.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
3K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
891
  • Introductory Physics Homework Help
Replies
3
Views
708
  • Introductory Physics Homework Help
Replies
9
Views
325
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top