A ball is dropped from the height of 50m. At a distance of 40m from the foot of perpendicular point p'is
observing the ball. Calculate the angular velocity (in rad/sec) of the point 'P' at t = 2sec.
Answers
Answer:
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Explanation:
Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which ω changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration α is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:
α
=
Δ
ω
Δ
t
α=ΔωΔt,
where Δω is the change in angular velocity and Δt is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s2. If ω increases, then α is positive. If ω decreases, then α is negative.
EXAMPLE 1. CALCULATING THE ANGULAR ACCELERATION AND DECELERATION OF A BIKE WHEEL
Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of -87.3 rad/s2, how long does it take the wheel to stop?
Strategy for (a)
The angular acceleration can be found directly from its definition in
α
=
Δ
ω
Δ
t
α=ΔωΔt because the final angular velocity and time are given. We see that Δω is 250 rpm and Δt is 5.00 s.
Solution for (a)
Entering known information into the definition of angular acceleration, we get
α
=
Δ
ω
Δ
t
=
250 rpm
5.00 s
.
α=ΔωΔt=250 rpm5.00 s.
Because Δω is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert Δω from rpm to rad/s:
Δ
ω
=
250
rev
min
⋅
2
π
rad
rev
⋅
1
min
60
sec
=
26.2
rad
s
Δω=250revmin⋅2π radrev⋅1 min60 sec=26.2rads
Entering this quantity into the expression for α, we get
α
=
Δ
ω
Δ
t
=
26.2 rad/s
5.00 s
=
5.24
rad/s
2
.
α=ΔωΔt=26.2 rad/s5.00 s=5.24 rad/s2.
Strategy for (b)
In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δt, yielding
Δ
t
=
Δ
ω
α
Δt=Δωα.
Solution for (b)
Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Δω is –26.2 rad/s, and α is given to be -87.3 rad/s2. Thus,
Δ
t
=
−
26.2 rad/s
−
87.3
rad/s
2
=
0.300 s.
Δt=−26.2 rad/s−87.3rad/s2=0.300 s.
Discussion
Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval.