Question

5.
A particle moves in such a way that time dependent x' and 'y' co-ordinates are given by the equations
x=ὦRt + R sin ὦt
y= R + R cos ὦt
Choose the correct alternative(s) from the following
(A) at the maximum value of its Y co-ordinate its speed is 2R
(B) at the minimum value of its Y co-ordinate its speed is zero
(C) at the maximum value of its Y co-ordinate its accelerations is equal to ὦ²R directed along negative y-axis.
(D) at the minimum value of its Y co-ordinate its accelerations is equal to ὦ²R directed along positive y-axis.

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Answer 1

Answer:

Explanation:

As  

x

=

a

sin

ω

t

and  

y

=

a

cos

ω

t

sin

ω

t

=

x

a

and  

cos

ω

t

=

y

a

Hence squaring and adding

sin

2

ω

t

+

cos

2

ω

t

=

x

2

a

2

+

y

2

a

2

or  

x

2

a

2

+

y

2

a

2

=

1

or  

x

2

+

y

2

=

a

2

Hence, it is (2) A circular path.

Explanation:

Answer 2

Answer:

B, C and D are correct options.

Explanation:

Given a particle moves where the time dependent x and y co-ordinates are given by the equations

$~\mbox{x}=\omega Rt + R ~\mbox{sin} \omega t ~\mbox{and}~y= R + R ~\mbox{cos} \omega t}$

The displacement vector can be written as

$\vec r=\big(\omega Rt + R ~\mbox{sin} \omega t\big)\hat i + \big(R + R ~\mbox{cos} \omega t}\big)\hat j$

Velocity is given by

$\vec v=\dfrac{d\vec r}{dt}$

  $=\dfrac{d}{dt}\Big[\big(\omega Rt + R ~\mbox{sin} \omega t\big)\hat i + \big(R + R ~\mbox{cos} \omega t}\big)\hat j\Big]$

  $=\big(\omega R + R \omega~\mbox{cos} \omega t\big)\hat i + \big(- R\omega ~\mbox{sin} \omega t}\big)\hat j\Big]$                              ...(1)

Acceleration is given by

$\vec a=\dfrac{d\vec v}{dt}$

  $=\dfrac{d}{dt}\Big[\big(\omega R + R \omega~\mbox{cos} \omega t\big)\hat i + \big(- R\omega ~\mbox{sin} \omega t}\big)\hat j\Big]}$

  $=\big(- R \omega^2~\mbox{sin} \omega t\big)\hat i + \big(- R\omega^2 ~\mbox{cos} \omega t}\big)\hat j$                                 ...(2)

(A) For maximum value of y co-ordinate,

we should have $\mbox{cos} \omega t=+1 ~\mbox{and}~\mbox{sin} \omega t=0$

Substituting these values in equation (1), at maximum y, the velocity is

$\vec v=\big(\omega R + R \omega)\hat i=2R \omega\hat i$

Therefore, statement A is incorrect.

(B) For minimum value of y co-ordinate,

we should have $\mbox{cos} \omega t=-1 ~\mbox{and}~\mbox{sin} \omega t=0$

Substituting these values in equation (1), at minimum y, the velocity is

$\vec v=\big(\omega R - R \omega)\hat i-0\hat j=0$

Therefore, statement B is correct.

(C) For maximum value of y co-ordinate, i.e., $\mbox{cos} \omega t=+1 ~\mbox{and}~\mbox{sin} \omega t=0$

Substituting these values in equation (2),

the acceleration at maximum y is

$\vec a=-\omega^2 R\hat j$

The acceleration is reducing, and the magnitude is equal to $\omega^2R$ directed along negative y-axis.

Therefore, statement C is correct.

(D) For minimum value of y, $\mbox{cos} \omega t=-1 ~\mbox{and}~\mbox{sin} \omega t=0$

Substituting these values in equation (2), the acceleration at minimum y is  

$\vec a=\omega^2 R\hat j$

The acceleration is equal to $\omega^2R$ directed along positive y-axis.

Therefore, statement D is correct.

Therefore, B, C and D are correct options.

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