3 Circular Motion Problems
1. A circular disk of radius 30 cm is rotating with an angular acceleration of 1 radian/ s2. What is the tangential acceleration of a point on the rim of the disk? What is its centripetal acceleration if the angular speed is 3 rev / s?
2. A 980 Kg car traveling at 20 m/s rounds a curve of radius 40 m. What is the friction force that must act on the car to keep it in its circular path?
3. An object with a mass of 100 Kg is dragged up an incline by applying a force F. The total distance moved along the incline is 200 m and the vertical height gained in moving this distance is 20 m. What is the magnitude of the force F if friction is negligible? (Assume g = 10 m/s2)
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Solution Preview
Please. find the solution to all the problems on circular motion in the attached file 'circular_motion_problems_001.doc'.
Note: If you don't know the definition of either angular velocity, tangential acceleration, centripetal acceleration etc. then you can read these to the conceptual level at the end of these problems. Try to understand all the figures conceptually. You will be able to solve most of the parallel problems on your own. For this, I you will find here at the end, the table of formulae included in this document.
All the best.
1. A circular disk of radius 30 cm is rotating with an angular acceleration of 1 radian/ s2. What is the tangential acceleration of a point on the rim of the disk? What is its centripetal acceleration if the angular speed is 3 rev / s?
Answer:
Data: radius r= 30 cm = 0.3 m (meter)
Angular Acceleration = 1 rad/s2
The tangential acceleration of a point on the rim of the disc at = ?
Tangential acceleration = radius x angular acceleration.
i.e. at = r x
i.e. at = 0.3 x 1
at = 0.3 m/s2
Centripetal Acceleration:
Given angular speed = 3 rev/s i.e. frequency of revolution, n = 3 rev/s.
Angular speed () = 2 x n
i.e. = (2) (3.14) (3) = 18.84 rad/s.
Centripetal acceleration is given by ac = V2/r but v = r
ac = r2 (r  radius, angular speed)
= 0.3 x (18.84)2 = 106.48 m/s2
ac = 106.48 m/s2

2. A 980 Kg car traveling at 20 m/s rounds a curve of radius 40 m. What is the friction force that must act on the car to keep it in its circular path?
Answer:
Data: m = 980 kg v = 20 m/s radius r = 40 m
Frictional force required to keep the car in the circular path F = ?
For the motion of the car in the circular path (on a horizontal road), the frictional force must at least be of same magnitude as that of the centripetal force acting on the car.
If Frictional force > Centripetal force  Car is safe.
Frictional force = Centripetal force  Car is safe (but on the verge of
danger.
Frictional force < Centripetal force  Car is not safe, it will be thrown
off the roads.
Thus for the safety of the car,
Frictional force = Centripetal force =

3. An object with a mass of 100 Kg is dragged up an incline by applying a force F. The total distance moved along the incline is 200 m and the vertical height gained in moving this distance is 20 m. What is the magnitude of the force F if friction is negligible? (Assume g = 10 m/s2).
Answer:
Data:
In fig. 1, AC is the inclined plane. dist. AC = 200 m.
The height of the end of the plane from the ground i.e. dist. BC = 20 m
Mass of the object m = 200 kg.
g = 10 m/s2
Magnitude of the force that is applied to the object to take it to the height h, F =?
As friction is to be assumed negligible, the only force acting on the object is its own weight, which acts vertically downward as shown in the figure. Refer figure (1).
This force F as seen from figure (2), can be resolved into its two perpendicular components .
(i) mg cos: Acting along the direction perpendicular to the inclined plane.
(ii) mg sin: along the plane, directed in the opposite direction of the
displacement of the object, as shown in figure (1).
The block must be displaced from point A to the point B against the force mg sin. This much amount of force must be applied to the block in the forward direction displace it, towards the point B.
From figure,
Thus the magnitude of the applied force is given by
F = mg sin
= 100 x 10 x 0.2 = 200 N

Try understanding the Relation between the tangential acceleration and the angular acceleration:
Relation between tangential acceleration (at) and angular acceleration ():
Consider a particle moving along a circular path of radius r as shown in figure.
Let r  radius of the circular path.
 Angular velocity of the particle at point A at time t.
t  time taken by the particle to travel from point A to point B.
small change in angular velocity of the particle in time t while going from point A to point B.
v  Change in magnitude of linear velocity of the particle in time t.
 Instantaneous tangential linear acceleration of the particle.
 (1)
But v = r .
By definition,
angular acceleration.
[ scalar ...
Solution Summary
13 page file which contains the solution to the problems plus the concepts of circular motion, with diagrams, table of formulae and few important derivations.
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