A particle -m- starts with zero velocity along a line y -4 d- Where d is a constant- The position of particle -m- varies as x - A sin- t- At - t -2- its angular momentum with respect to the origin is-A- m A - dB- m A - d-AD- zero

The correct option is D zeroAt ωt=π/2 u=dx/dt=Aω cos ωt=Aω cos ( π/2 )=0 So the angular momentum about origin = 0. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Cross Product

A particle 'm...

Question

A particle 'm' starts with zero velocity along a line y = 4d. (Where d is a constant). The position of particle 'm' varies as $x = A s i n ω t . A t ω t = \frac{π}{2}$ , its angular momentum with respect to the origin is?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D zero
$A t ω t = \frac{π}{2} u = \frac{d x}{d t} = A ω c o s ω t = A ω c o s (\frac{π}{2}) = 0$
So the angular momentum about origin = 0.

Suggest Corrections

Similar questions

A particle ' $m$ ' starts with zero velocity along a line $y$ = $4 d$ . (Where $d$ is a constant). The position of particle ' $m$ ' varies as $x$ = $A s i n ω t$ . At $ω t$ = $\frac{π}{2}$ , its angular momentum with respect to the origin is?

Q. The velocity of a particle of mass

m

moving along a straight line varies with time

t

\frac{d^{2} v}{d t^{2}} = - k v

, the particle's acceleration was zero when it started its motion from

x = 0

. Here

k

is a positive constant. Choose the incorrect statement.

Q. The position x of a particle moving along x-axis varies with time t as x=Again(ωt) where A and ω are positive constants. The acceleration of a particle varies with its position (x) as

Q. The position x of particle moving along x- axis varies with time t as x= A sin

(ω t)

where A and

ω

are positive constants. The acceleration a of particle varies with its position (x) as

Q. A particle located at

x = 0

at time

t = 0

, starts moving along the positive

x -

direction with a velocity

v

that varies with position as

v = a \sqrt{x}

. Where

a

is constant. Then the displacement of the particle varies with time as:

Join BYJU'S Learning Program

A particle 'm' starts with zero velocity along a line y = 4d. (Where d is a constant). The position of particle 'm' varies as x=A sin ωt. At ωt=π2 , its angular momentum with respect to the origin is?

The correct option is D zeroAt ωt=π2u=dxdt=Aω cos ωt=Aω cos(π2)=0 So the angular momentum about origin = 0.

A particle 'm' starts with zero velocity along a line y = 4d. (Where d is a constant). The position of particle 'm' varies as $x = A s i n ω t . A t ω t = \frac{π}{2}$ , its angular momentum with respect to the origin is?

The correct option is D zero
$A t ω t = \frac{π}{2} u = \frac{d x}{d t} = A ω c o s ω t = A ω c o s (\frac{π}{2}) = 0$
So the angular momentum about origin = 0.