Question

A particle moves in xy plane on a circular path of radius
10 m (with centre at origin) with constant speed of 20 m/s.
Its velocity vector at point (0,10 m) if it's angular velocity
is along z-axis is
(1) (20i)m/s (2) (-20i) m/s
(3) (10i)m/s (4) (-10i)m/s​

Answer 1

Change in velocity of a particle in circle in given by,  

∣Δ  

v

ˉ

∣=  

v  

1

2

​  

+v  

2

2

​  

+2 v  

1

​  

v  

2

​  

cos ( π−i )

​  

 

Since [∣  

v  

1

​  

 

ˉ

​  

∣=∣  

v  

2

​  

 

ˉ

​  

∣]=v  

∣Δ  

v

ˉ

∣=2 vwithout  

2

θ

​  

 

On putting the values of vandθ

=(2×10)×sin ( 3 0  

The

)

=10m/s

Hence the change in velocity is 10m/s

Answer 2

A particle moves in xy plane on a circular path of radius 10 m ( with centre at origin) with constant speed of 20 m/s.

We have to find the velocity vector at a point ( 0, 10) if it's angular velocity is along z- axis.

method 1 : in this method, you can find the unit vector of velocity vector using a standard formula of circular motion. let's find it.

position vector, $\vec r=10\hat j$ [ you see, only y axis of the point has given. ]

so the unit vector along position vector is $\hat j$

a/c to question, the angular velocity is along z-axis. so the unit vector of angular velocity along z-axis is $\hat k$ .

we know, for circular motion, $\vec{v}=\vec{\omega}\times\vec{r}$

so the unit vector along the velocity vector is , $\hat v=\hat k\times\hat j=-\hat i$

therefore the velocity vector would be , $\vec v=|v|\hat v$ = -20i

method 2 : using right hand rule of cross product, we can find out the velocity vector of the particle. if we wrap out our fingers in the direction of turning then the direction of angular velocity which is shown by the thumb of right hand is along z - axis.

see diagram as shown in figure.

so, the velocity vector of the particle, $\vec v=-20\hat i$