Advertisements
Advertisements
प्रश्न
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown figure. If he maintains constant speed of 10 ms–1, what is his acceleration at point R in magnitude and direction?
Advertisements
उत्तर
According to the problem, the path of the cyclist is OPRQO.
The cyclist is in uniform circular motion and it is given that linear velocity = 10 m/s, R = 1 km = 1000 m. As we know whenever an object is performing a circular motion, acceleration is called centripetal acceleration and is always directed towards the centre. So cyclist experiences a centripetal force (acceleration) at point R towards the centre.
The centripetal acceleration at R is given by the relation, `a_c = v^2/r`
⇒ `a_c = (10)^2/1000`
= `100/10^3`
= 0.1 m/s2 along RO.
APPEARS IN
संबंधित प्रश्न
Is it possible to have an accelerated motion with a constant speed? Name such type of motion.
A piece of stone tied at the end of a thread is whirled in a horizontal circle with uniform speed by hand. Answer the following questions:
- Is the velocity of stone uniform or variable?
- Is the acceleration of stone uniform or variable?
- What is the direction of acceleration of stone at any instant?
- Which force provides the centripetal force required for circular motion?
- Name the force and its direction which acts on the hand.
The motion of the moon around the earth in a circular path is an accelerated motion.
A uniform linear motion is unaccelerated, while a uniform circular motion is an accelerated motion.
Define angular velocity.
Solve the following problem.
A car moves in a circle at a constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of the acceleration of the car.
Solve the following problem.
A particle moves in a circle with a constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
The angle subtended by the vector A = `5hat"i" + 3hat"j" + 12hat"k"` with the X-axis is ______.
If a particle moves with uniform speed then its tangential acceleration will be ______.
The ratio of the angular speed of minute hand and hour hand of a watch is ____________.
A string of length 'l' fixed at one end carries a mass 'm' at the other end. The string makes `3/pi` revolutions/second around the vertical axis through the fixed end as shown in figure. The tension 'T' in the string is ______.
For a particle performing uniform circular motion, choose the correct statement(s) from the following:
- Magnitude of particle velocity (speed) remains constant.
- Particle velocity remains directed perpendicular to radius vector.
- Direction of acceleration keeps changing as particle moves.
- Angular momentum is constant in magnitude but direction keeps changing.
Earth also moves in circular orbit around sun once every year with on orbital radius of 1.5 × 1011 m. What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
A particle is given an initial speed u inside a smooth spherical shell of radius R = 1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is ______.
A disc of radius 5 cm rolls on a horizontal surface with linear velocity v = 1`hat"i"` m/s and angular velocity 50 rad/s. Height of particle from ground on rim of disc which has velocity in vertical direction is ______ cm.
A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is ______.
In uniform circular motion, although the speed is constant, why does acceleration occur?
