0 95s Round 2 : 338 Q18 of 30 English A boy runs on a circular road with a uniform speed of a 15 m/s. The radius of the path is 20 m. What will be the centripetal acceleration and its direction?.
Answers
Answer:
The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r and Δs are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v1 = v2 = v. Using the properties of two similar triangles, we obtain
Δ
v
v
=
Δ
s
r
.
Acceleration is
Δ
v
Δ
t
, and so we first solve this expression for Δv:
Δ
v
=
v
r
Δ
s
.
Then we divide this by Δt, yielding
Δ
v
Δ
t
=
v
r
×
Δ
s
Δ
t
.
Finally, noting that
Δ
v
Δ
t
=
a
c
and that
Δ
s
Δ
t
=
v
, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is
a
c
=
v
2
r
,
which is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that ac is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that ac is greater for tighter turns, as you have probably noticed.
It is also useful to express ac in terms of angular velocity. Substituting v = rω into the above expression, we find
a
c
=
(
r
ω
)
2
r
=
r
ω
2
. We can express the magnitude of centripetal acceleration using either of two equations:
a
c
=
v
2
r
;
a
c
=
r
ω
2
.
Recall that the direction of ac is toward the center. You may use whichever expression is more convenient, as illustrated in examples below.
Answer: