Question

A particle starting from rest moves with constant angular acceleration alpha in a circular path the time at which magnitude of tangential and radial acceleration are equal

Answer 1

Answer:

Explanation:

As we realize that  

Centripetal speeding up = v^2/r  

To start with condition of movement  

ω = ωo+ αt  

As beginning rakish speed is 0  

ω = αt  

v/r = 4t  

v = 4tr.  

Extraneous increasing speed = centripetal quickening  

αr = a  

4r = v^2/r  

4r = (4tr)2/r  

4 = 16t^2  

t = ½  

t = 0.5 sec.

Cheers!!

Regards,

Vikas (B. Tech. 4th year

Thapar University)

Answer 2

this is the answer option (c) is correct answer