From this equation we see that the acceleration vector has magnitude
$A{\omega}^{2}$ and is directed opposite the position vector, toward the origin, because
$\overrightarrow{a}(t)=\text{\u2212}{\omega}^{2}\overrightarrow{r}(t).$
Circular motion of a proton
A proton has speed
$5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and is moving in a circle in the
xy plane of radius
r = 0.175 m. What is its position in the
xy plane at time
$t=2.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\phantom{\rule{0.2em}{0ex}}\text{s}=200\phantom{\rule{0.2em}{0ex}}\text{ns?}$ At
t = 0, the position of the proton is
$0.175\phantom{\rule{0.2em}{0ex}}\text{m}\widehat{i}$ and it circles counterclockwise. Sketch the trajectory.
Solution
From the given data, the proton has period and angular frequency:
The position of the particle at
$t=2.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\phantom{\rule{0.2em}{0ex}}\text{s}$ with
A = 0.175 m is
From this result we see that the proton is located slightly below the
x -axis. This is shown in
[link] .
Significance
We picked the initial position of the particle to be on the
x- axis. This was completely arbitrary. If a different starting position were given, we would have a different final position at
t = 200 ns.
Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showing an acceleration in the direction of the motion.
In uniform circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. If the speed of the particle is changing as well, then we introduce an additional acceleration in the direction tangential to the circle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In
Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocity vector. If the speed of the particle is changing, then it has a
tangential acceleration that is the time rate of change of the magnitude of the velocity:
The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially inward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a
total acceleration that is the vector sum of the centripetal and tangential accelerations:
The acceleration vectors are shown in
[link] . Note that the two acceleration vectors
${\overrightarrow{a}}_{\text{C}}$ and
${\overrightarrow{a}}_{\text{T}}$ are perpendicular to each other, with
${\overrightarrow{a}}_{\text{C}}$ in the radial direction and
${\overrightarrow{a}}_{\text{T}}$ in the tangential direction. The total acceleration
$\overrightarrow{a}$ points at an angle between
${\overrightarrow{a}}_{\text{C}}$ and
${\overrightarrow{a}}_{\text{T}}.$
Questions & Answers
Can any one give me the definition for Bending moment plz...
A computer is reading from a CD-ROM that rotates at 780 revolutions per minute.What is the centripetal acceleration at a point that is 0.030m from the center of the disc?
the specific heat of hydrogen at constant pressure and temperature is 14.16kj|k.if 0.8kg of hydrogen is heated from 55 degree Celsius to 80 degree Celsius of a constant pressure. find the external work done .
Many amusement parks have rides that make vertical loops like the one shown below. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of 3.7 m/s2 .
At a post office, a parcel that is a 20.0-kg box slides down a ramp inclined at 30.0° 30.0° with the horizontal. The coefficient of kinetic friction between the box and plane is 0.0300. (a) Find the acceleration of the box. (b) Find the velocity of the box as it reaches the end of the plane, if the length of the plane is 2 m and the box starts at rest.