Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
saysaban
Group Title
Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say rn, and some power of v, say vm. Determine the values of n and m and write the simplest form of an equation for the acceleration.
 9 months ago
 9 months ago
saysaban Group Title
Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say rn, and some power of v, say vm. Determine the values of n and m and write the simplest form of an equation for the acceleration.
 9 months ago
 9 months ago

This Question is Open

adekurniawan Group TitleBest ResponseYou've already chosen the best response.1
the value of m and n can be obtained using a tool called dimensional analysis acceleration a has dimension\[L T^{2}\] the velocity has dimension \[LT ^{1}\] and radius of a circle (or any path) is L, the equation of acceleration, can be written \[a = v^{m}r^{n}\]. put the dimensions of a, v, r in the equation respectively yields \[LT ^{2}=\left( LT ^{1} \right)^{m}\left( L \right)^{n}\]. Simplyfing the results, \[LT ^{2}=L^{m+n}T^{m}\]. By comparing the power of L and T on the left hand with the right hand one, we have\[m=2\] and \[m+n=1\]. Substitung m=2 into the last equation,we obtain n=1.And finally we have equation of a in terms of v and r by substituting m and n\[a=v^{m}r ^{n}\] \[a=v ^{2}r ^{1}\] or \[a=\frac{v ^{2} }{ r }\] which is the wellknown centripetal acceleration
 9 months ago

saysaban Group TitleBest ResponseYou've already chosen the best response.0
thank you
 9 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.