Question

A point moves along a circle having a radius 20cm with a constant tangential acceleration 5cm//s^(2) . How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration?

Answer 1

The answer is 2 seconds.

The point moves along a circle having a radius 20  cm with a constant tangential acceleration 5 cm/s² .

• Tangential acceleration is given by :  a(t) = dv/dt
• Normal acceleration is given by : a(n) =  v²/r

1. a(t) = 5 cm/s²
2. After time t, a(n) = a(t)
3. a(n) = 5 cm/s²
4. $\frac{v^{2} }{r} = 5$
5. v = $\sqrt{5 . 20}$
6. v = 10 cm/s

• Now, dv/dt = 5

1. dv = 5 dt
2. v = 5 t
3. t = 10/5

 t = 2 seconds

             

Answer 2

GIVEN :

A point moves along a circle with radius 20cm, tangential acceleration 5cm//s^(2) .

TO FIND :

Time needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration.

SOLUTION:

◆Tangential acceleration ,

 a= dv / dt

◆Normal acceleration ,

a' =  v²/r

◆ It is given that,

a= 5 cm/s²

After time t, a' = a

◆Thus,

a'= 5 cm/s²

v^2/r = 5 cm/s²

v = √(5 ×20)

v = 10 cm/s.

◆Here,

a = dv/dt = 5

dv = 5 dt

0-v = ( 0-5 )t ;

(as the motion starts from rest)

◆t = 10/5 = 2s

ANSWER :

Time needed after motion begins from normal acceleration of the point to be equal to tangential acceleration is 2s.