What will be the anser?

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2 Answers
Aug 11, 2018

5.7 m/s5.7ms

Explanation:

Here,considering speed to be the magnitude of linear velocity.

So,if it is moving with constant speed of vv at a certain point of time,then its Centripetal acceleration is |vec a_c|=v^2/rac=v2r where rr is the radius of the circular path.

If its linear acceleration is vec a_rar then,

vec a_c + vec a_r =vec aac+ar=a

Given, |vec a|=15 ms^-2a=15ms2

Again angle between vec a_cac and vec a_rar is 90^@90

So, a^2=a_r^2 + a_c^2a2=a2r+a2c

Or, a^2=a_r^2 + v^4/r^2... 1

Given, a=15 ms^-2,r=2.5m

If, vec a makes an angle of 30^@ w.r.t vec a_c then,

tan 30=|vec a_r| /|vec a_c|

I.e 1/sqrt(3) =a_r/(v^2/r)

Given, r=2.5m

So, putting the values and arranging we get,

a_r =v^2/(2.5 sqrt(3))

Putting this value of a_r in the previous equation 1 we get,

v=5.7 m/s

Aug 11, 2018

Slightly shorter approach, same answer

Acceleration vector in non-uniform circular motion is:

  • bba(t) = underbrace(r dot theta \ bbhat e_theta)_("tangential ") - underbrace(v^2/r\ bb hat e_r)_("inward radial: " = bba_r)

In given geometry:

abs(bba_r) = abs(bba ) cos 30

implies v^2 /2.5 = 15 sqrt3/2

:. "speed" = abs(bbv) ~~ 5.7 " m/s"