What is the computation of tangential acceleration in circular motion?

AI Thread Summary
Tangential acceleration in circular motion is calculated using the formula that relates it to the rate of change of speed, which is given as 0.600 m/s² in this case. The automobile travels along a circular path with a radius of 20.0 m and an instantaneous speed of 3.00 m/s. While tangential acceleration affects the speed, centripetal acceleration, which is perpendicular to it, maintains the circular motion. The total acceleration can be viewed as the hypotenuse of a right triangle formed by tangential and centripetal accelerations. Understanding these components is essential for solving problems related to circular motion.
lostinphys
Messages
8
Reaction score
0
I'm sorry guys... I'm stuck again...
An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 3.00 m/s, find the tangential acceleration component.
Magnitude.

... i know that the tangential acceleration = radius (angular acceleration) , but i am having trouble understanding the actual computation of tangential acceleration. i picked up a couple of books, and I'm still pretty lost.
 
Physics news on Phys.org
lostinphys said:
I'm sorry guys... I'm stuck again...
An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 3.00 m/s, find the tangential acceleration component.
Magnitude.

... i know that the tangential acceleration = radius (angular acceleration) , but i am having trouble understanding the actual computation of tangential acceleration. i picked up a couple of books, and I'm still pretty lost.


hi there...

im new to this site.. sorry i can't help you w/ that problem... maybe you can help me with a simple question i have?
 
Centripetal acceleration does not affect speed (it changes the direction of velocity).
I'd say the tangental acceleration is 0,6 m/s^2.


If 0,6 m/s^2 was, however, meant as the total acceleration:
If the instantaneous velocity at the given instant t is 3,00 m/s, what is the centripetal acceleration needed to keep the object on its circular orbit of 20m radius?
Tangental and centripetal acceleration are perpendicular. The given total acceleration is the hypotenuse.
 
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