Question

A particle moves in a circle of radius 1.0cm with a speed given by v=2t , where v is in cm//s and t in seconds. (a) Find the radial accerleration of the particle at t=1s . (b) Find the tangential accerleration of the particle at t=1s . Find the magnitude of net accerleration of the particle at t=1s .

Answer 1

Answer:

Radius, r = 1.0 cm

(a.) t = 1s

Velocity of the particle, u = 2.0 t = 2.0×1 = 2.0 cms-1

Radial acceleration of the particle = v2/r= (2.0)2/10 = 4.0 cm/s2

(b.) It is given that v=2t

Differentiating both sides with respect to time,

⇒ dv/dt =2 dt/dt

⇒ a=2 cms-2

⇒ Tangential acceleration, a = 2 cm/s2

(c.) Magnitude of acceleration

=

$\sqrt{4 {}^{2} } + 2 {}^{2}$

=√20cm/s2

Answer 2

Radial acceleration = 4 cm/s²

Tangential acceleration = 2 cm/s²

Net acceleration = 4.47  cm/s²

Given:

A particle moves in a circle of radius 1.0 cm with a speed given by v=2t,  v is in cm/s.

To find:

(a) The radial acceleration of the particle at t=1s .

(b) The tangential acceleration of the particle at t=1s .

(c) The magnitude of net acceleration of the particle at t=1s .

Explanation:

For t = 1 sec.    v=2t = 2×1 = 2 cm/s.

(a) The radial acceleration of the particle at t = 1 sec.

  a = $\frac{V^2}{r}$ = $\frac{2^2}{1}$ = 4 cm/s²

(b) The tangential acceleration of the particle at t=1s.

  a = $\frac{dv}{dt}$ =$\frac{d}{dt}$ (2t) = 2 cm/s²

(c) The magnitude of net acceleration of the particle at t=1s.

 Net acceleration = $\sqrt{4^2+2^{2} }$ = 4.47  cm/s²

To learn more...

1)The length of hour hand of a wrist watch is 1.5 cm. Find magnitude radial acceleration.

(Ans : 3.175 x 10⁻¹⁰ m/s²)

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