if a body is moving in circular path of radius 14 m.it complete one resolution in 7 sec considering the speed to be uniform the value of centripital acceleration is
Answers
Answer:
Explanation:
Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.
The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration a_ca
c
a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.
The directions of the velocity of an object at two different points, \text BBstart text, B, end text and \text CCstart text, C, end text, are shown, and the change in velocity, \Delta vΔvdelta, v, is seen to point roughly toward the center of curvature. To see what happens instantaneously, the points \text BBstart text, B, end text and \text CCstart text, C, end text must be very close and \Delta \thetaΔθdelta, theta very small. We'll then find that \Delta vΔvdelta, v points directly toward the center of curvature.
Because a_c=\dfrac{\Delta v}{\Delta t}a
c
=
Δt
Δv
a, start subscript, c, end subscript, equals, start fraction, delta, v, divided by, delta, t, end fraction, the acceleration is also toward the center. Because \Delta \thetaΔθdelta, theta is very small, the arc length \Delta sΔsdelta, s is equal to the chord length \Delta rΔrdelta, r for small time differences. Image credit: Openstax College Physics
The direction of centripetal acceleration is toward the center of the circle, but what is its magnitude? Note that the triangle formed by the velocity vectors and the triangle formed by the radii rrr and \Delta sΔsdelta, s are similar. Both the triangles ABCABCA, B, C and PQRPQRP, Q, R are isosceles triangles with two equal sides. The two equal sides of the velocity vector triangle are the speeds v_1=v_2=vv
1
=v
2
=vv, start subscript, 1, end subscript, equals, v, start subscript, 2, end subscript, equals, v. Using the properties of two similar triangles, we obtain \dfrac{\Delta v}{v}=\dfrac{\Delta s}{r}
v
Δv
=
r
Δs
start fraction, delta, v, divided by, v, end fraction, equals, start fraction, delta, s, divided by, r, end fraction.
Acceleration is \dfrac{\Delta v}{\Delta t}
Δt
Δv
start fraction, delta, v, divided by, delta, t, end fraction, so we first solve the above expression for \Delta vΔvdelta, v:
\Delta v=\dfrac{v}{r}\Delta sΔv=
r
v
Δsdelta, v, equals, start fraction, v, divided by, r, end fraction, delta, s
If we divide both sides by \Delta tΔtdelta, t we get the following:
\dfrac{\Delta v}{\Delta t}=\dfrac{v}{r}\times \dfrac{\Delta s}{\Delta t}
Δt
Δv
=
r
v
×
Δt
Δs
start fraction, delta, v, divided by, delta, t, end fraction, equals, start fraction, v, divided by, r, end fraction, times, start fraction, delta, s, divided by, delta, t, end fraction
Finally, noting that \dfrac{\Delta v}{\Delta t}=a_c
Δt
Δv
=a
c
start fraction, delta, v, divided by, delta, t, end fraction, equals, a, start subscript, c, end subscript and that \dfrac{\Delta s}{\Delta t}=v
Δt
Δs
=vstart fraction, delta, s, divided by, delta, t, end fraction, equals, v, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is a_c=\dfrac{v^2}{r}a
c
=
r
v
2
a, start subscript, c, end subscript, equals, start fraction, v, squared, divided by, r, end fraction.
This is the acceleration of an object in a circle of radius rrr at a speed vvv. So, centripetal acceleration is greater at high speeds and in sharp curves—smaller radii—as you have noticed when driving a car. But it is a bit surprising that a_ca
c
a, start subscript, c, end subscript is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/hr than at 50 km/hr. A sharp corner has a small radius, so a_ca
c
a, start subscript, c, end subscript is greater for tighter turns, as you have probably noticed.
Answer:
78.8768m/s
Explanation:
we know that
speed = distance
-------------
time
i.e
v=x/t
;v=2πr/7
v=2π(14)/7
v=4π.
Now
centripital acceleration=v²/r
=(4π)²/7
=78.876m/s