Angular Speed of Electric Toothbrush Head: 795.49rad/s

  • Thread starter Thread starter bigtymer8700
  • Start date Start date
AI Thread Summary
The discussion focuses on calculating the average angular speed of an electric toothbrush head that rotates back and forth through a 70-degree angle at a rate of 7600 times per minute. The initial calculation of 795.49 rad/s is incorrect because it assumes a full rotation instead of the specified angle. The correct approach involves converting the 70 degrees to radians and considering the time for each back-and-forth cycle. If the toothbrush head rotates 70 degrees in one direction and then returns, the angular speed would effectively be doubled. The clarification emphasizes the importance of understanding the motion's direction and the angle involved in the calculation.
bigtymer8700
Messages
40
Reaction score
0
Electric toothbrushes can be effective in removing dental plaque. One model consists of a head 1.10cm in diameter that rotates back and forth through a 70.0 angle 7600times/min . The rim of the head contains a thin row of bristles. its asking me the average angular speed in each direction of the rotating head.

w=7600times/min(1min/60s)6.28rad/1rev= 795.49rad/s

but it says I am wrong what am i doing wrong?
 
Last edited:
Physics news on Phys.org
It rotates through 70deg 7600 times /minute not a full circle.
so that's 70 * pi/180 radians in 60/7600 seconds.

It could also be double that if it means it goes +70deg then back to the start 7600 times a minute?
 
yea it would be double since its going back and forth thanks
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top