
A vehicle is moving due north with an absolute velocity 54 km/h. An observer at P is at distance 30 m to the west of the line of travel. What is the angular velocity of the vehicle relative to the observer at t = 0 and t = 2 s.
Answer
533.4k+ views
Hint: Angular velocity is the ratio of linear velocity and radius or the distance from the centre point. The distance from the centre is changing when the vehicle is moving in the straight line towards north.
Formula used:
The angular velocity and the linear velocity are related as:
$v = \omega r$
Complete answer:
The vehicle is moving due north and the given point P is stationary point towards west. As the vehicle is present at the origin, the perpendicular distance is 30 m, here the time t = 0s, we need to find the distance by which the vehicle travels due north with 54 km/h velocity. The distance covered is the product of linear velocity and time:
$d = \dfrac{54 \times 1000}{3600} \times 2 = 30$ m.
Therefore, after 2s, the car moves a distance of 30 m towards north as it starts from the origin. The distance from the point P has to be determined by using Pythagoras theorem in the following manner:
$r = \sqrt{30^2 + 30^2} = 30\sqrt{2}$ m.
Now, we have our two values that we will put in the formula for angular velocity. Therefore,
(1) at t = 0 s, the distance/radius PO is 30 m and the velocity v is 54 m/s. The angular velocity will be:
$\dfrac{54 \times 1000}{30 \times 3600 } = 0.5 $ rad/s.
(2) at t = 2 s, the distance/radius will become $30\sqrt{2}$ m. Now, in this case, the linear velocity is not exactly perpendicular to angular velocity and radius, so we take the projection of that linear velocity due north on the perpendicular direction.
The angular velocity becomes:
$\dfrac{54 \times 1000 \cos 45^\circ}{30\sqrt{2} \times 3600} = 0.25 $ rad/s.
Note:
Here, one can easily drop the fact that we are dealing with vector quantities and one might deal with everything in scalar terms. But one must remember that the linear velocity is the cross product of angular velocity and radius vector. The angular velocity points out of the page and towards us (right hand rule), radius vector is along r, then linear velocity has to be perpendicular to both.
Formula used:
The angular velocity and the linear velocity are related as:
$v = \omega r$
Complete answer:
The vehicle is moving due north and the given point P is stationary point towards west. As the vehicle is present at the origin, the perpendicular distance is 30 m, here the time t = 0s, we need to find the distance by which the vehicle travels due north with 54 km/h velocity. The distance covered is the product of linear velocity and time:
$d = \dfrac{54 \times 1000}{3600} \times 2 = 30$ m.
Therefore, after 2s, the car moves a distance of 30 m towards north as it starts from the origin. The distance from the point P has to be determined by using Pythagoras theorem in the following manner:
$r = \sqrt{30^2 + 30^2} = 30\sqrt{2}$ m.
Now, we have our two values that we will put in the formula for angular velocity. Therefore,
(1) at t = 0 s, the distance/radius PO is 30 m and the velocity v is 54 m/s. The angular velocity will be:
$\dfrac{54 \times 1000}{30 \times 3600 } = 0.5 $ rad/s.
(2) at t = 2 s, the distance/radius will become $30\sqrt{2}$ m. Now, in this case, the linear velocity is not exactly perpendicular to angular velocity and radius, so we take the projection of that linear velocity due north on the perpendicular direction.
The angular velocity becomes:
$\dfrac{54 \times 1000 \cos 45^\circ}{30\sqrt{2} \times 3600} = 0.25 $ rad/s.
Note:
Here, one can easily drop the fact that we are dealing with vector quantities and one might deal with everything in scalar terms. But one must remember that the linear velocity is the cross product of angular velocity and radius vector. The angular velocity points out of the page and towards us (right hand rule), radius vector is along r, then linear velocity has to be perpendicular to both.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

