Answer:
(a) 0.158 rad/s.
(b) 7.1 m/s² toward the center of the circular track.
Explanation:
(a)
Using,
v = ωr.................. Equation 1
Where v = speed of the car, r = radius of the circular track. ω = angular speed.
make ω the subject of the equation
ω = v/r ................... Equation 2
Given: v = 45 m/s, r = 285 m.
substitute into equation 2
ω = 45/285
ω = 0.158 rad/s
Hence the magnitude of the angular speed of the car = 0.158 rad/s.
(b)
Using
a = v²/r......................... Equation 3
Where a = acceleration of the car
Given: v = 45 m/s, r = 285 m
Substitute into equation 3
a = 45²/285
a = 2025/285
a = 7.1 m/s² toward the center of the circular track.
Answer:
A) Angular velocity = 0.1579 rad/s
B) Acceleration = 7.11 m/s² in a direction towards the centre of the circle motion
Explanation:
In circular motion,
linear speed = angular speed x radius of the rotation
Thus,
v = ωr
Where;
v = linear speed (m/s)
ω = angular speed (radians/s)
r = radius of the rotation (m)
We are looking for angular speed, thus ;making ω the subject, we have;
ω = v/r
We are given that v= 45 m/s and r = 285m
Thus,
ω = 45/285 = 0.1579 rad/s
B) The formula for Centripetal acceleration is given as;
a_c is given by the expression
a_c = rω² = v²/r
where;
v is linear velocity of the object,
ω is its angular velocity and
r is the radius of the circle in which object moves.
Thus,
a_c = v²/r = 45²/285 = 7.11 m/s²
The centripetal acceleration affects the direction of the car. So in this case the direction of the car will be towards the centre of the circle motion.
