A particle moving along the circular path with a speed v and its speed increases by g in one second. If the radius of the circular path be r, then the net acceleration of the particle is:

Question

A particle moving along the circular path with a speed v and its speed increases by g in one second. If the radius of the circular path be r, then the net acceleration of the particle is: (A) (v2/r)+g (B) (v2/r2)+g2 (C) ((v2/r)+g)1 / 2 (D) ((v4/r2)+g2)1 / 2

Tardigrade Admin · Accepted Answer

Net acceleration = √a2c +a2t In circular motion magnitude of velocity only changes by tangential acceleration. tangential acceleration = g ∴ a = √a2c +a2t a=√(v4/r2) +g2

Q. A particle moving along the circular path with a speed $v$ and its speed increases by $g$ in one second. If the radius of the circular path be $r$ , then the net acceleration of the particle is:

$\frac{v ^{2}}{r} + g$

$\frac{v ^{2}}{r ^{2}} + g^{2}$

$(\frac{v ^{2}}{r} + g)^{1/2}$

$(\frac{v ^{4}}{r ^{2}} + g^{2})^{1/2}$

Net acceleration $= a_{c}^{2} + a_{t}^{2}$
In circular motion magnitude of velocity only changes by tangential acceleration.
tangential acceleration $= g$
$∴ a = a_{c}^{2} + a_{t}^{2}$
$a = \frac{v ^{4}}{r ^{2}} + g^{2}$

Q. A particle moving along the circular path with a speed v and its speed increases by g in one second. If the radius of the circular path be r, then the net acceleration of the particle is:

rv2​+g

r2v2​+g2

(rv2​+g)1/2

(r2v4​+g2)1/2