Question

Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 6.00 m/s2? (b) can it have an acceleration of magnitude 4.00 m/s2? in each case, if the answer is yes, explain how it can happen; if the answer is no, explain why not.

Answer 1

Circular acceleration has 2 components, the centripetal and tangential acceleration.

The formula for calculating the centripetal acceleration of a particle is:

$a_{c}$ = v^2/r

Where v is the velocity or instantaneous speed and r is the radius.

$a_{c}$  = (3^2)/6 = 1.5 m/s^2

Assuming that the particle stays on the same circular path, another thing it can have is the tangential acceleration. While the centripetal acceleration is constant, tangential acceleration depends on the location of the particle in the circular path. So to answer A and B, yes we can have an acceleration of 6 and 4 respectively.

Answer 2

Answer:

(a) So yes acceleration can have a magnitude $6 \rm m/s^{2}$

(b)acceleration can not have a value $4 \rm m/s^{2}$ as it is lesser than $4.5 \rm m/s^{2}$

Explanation:

Given information:

Instantaneous speed ,$v=\rm 3.00\ m/s$

Radius of curvature of path $r=\rm 2\ m$

We know,

$\vec{a}_{total}= \vec{a}_c \ + \vec{ a}_t$

Where, $a_c$ is centripetal acceleration and $a_t$ is tangential acceleration

The centripetal acceleration of particle is

$a_c=\frac{v^{2} }{r} \rm m/s^{2}$=$\frac{3^{2} }{2} \rm ms^{-2}$

$a_c=9/2=4.5\rm ms^{-2}$

Magnitude of total acceleration is

$\left | a \right |=\sqrt{a_c^{2} +a_t^{2} }$

so

$\sqrt{a_c^{2} +a_t^{2} } \geq a_x$ , which means total acceleration can be greater than or equal to $4.5 \rm m/s^{2}$

(a) So yes acceleration can have a magnitude $6 \rm m/s^{2}$

(b)acceleration can not have a value $4 \rm m/s^{2}$ as it is lesser than $4.5 \rm m/s^{2}$.

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