13. doubles
The tension in the string corresponds to the centripetal force that holds the object in rotation, so:
[tex]T=F=m\frac{v^2}{r}[/tex]
where m is the mass of the object, v is the tangential speed, and r is the distance of the object from the centre of rotation (therefore it corresponds to the length of the string). The problem tells us that the tangential speed remains the same (v), while the length of the string is halved, so r'=r/2. Therefore, the new tension in the string will be
[tex]T'=m\frac{v^2}{r'}=m\frac{v^2}{r/2}=2m\frac{v^2}{r}=2T[/tex]
so, the Tension doubles.
14. Variations of centripetal forces
Both revolution and rotation refer to the rotational motion of an object, therefore they both involve the presence of a centripetal force, which keeps the object in circular motion. The only difference between the two is:
- Revolution is the circular motion of an object around a point external to the object (for instance, the motion of the Earth around the Sun)
- Rotation is the circular motion of an object around its centre, so around a point internal to the object (for instance, the rotation of the Earth around its axis)
15. Rotational speed
For a uniform object in circular motion, all the points of the object have same rotational speed. In fact, the rotational speed is defined as
[tex]\omega=\frac{\Delta \theta}{\Delta t}[/tex]
where [tex]\Delta \theta[/tex] is the angular displacement covered in a time interval of [tex]\Delta t[/tex]. Since all the points of the wheel are coeherent (they move together), they all cover the same angular displacement in the same time, so they all have same rotational speed.
16. away from the center of the path.
The tension in the string is responsible for keeping the tin can in circular motion. Therefore, the tension in the string represents the centripetal force, and so it is directed towards the centre of the path. According to Newton's third law, the tin can exerts a force on the string which is equal in magnitude (so, same magnitude of the tension), but opposite in direction: therefore, away from the centre of the path.
17. weight of the bob.
There are two forces acting on the bob in the vertical direction: the weight of the bob (downward) and the vertical component of the string tension (upward). Since there is no acceleration along the vertical direction, the net force must be zero, so these two forces must be equal: it means that the vertical component of the string tension is equal to the weight of the bob. Along the horizontal direction, instead, the horizontal component of the string tension corresponds to the centripetal force that keeps the bob in circular motion.
18. horizontal component of string tension.
Along the horizontal direction, there is only one force acting on the bob: the horizontal component of the string tension. Since the bob is moving of circular motion along the horizontal direction, this means that this force (the horizontal component of the string tension) must correspond to the centripetal force that keeps the pendulum in circular motion.
19. inward, toward the center of swing.
The force that the can exerts on the bug is the force that keeps the bug in circular motion (since it prevents the bug from moving away). Therefore, it must corresponds to the centripetal force.
20. speed of the car. AND radius of curvature.
The normal force exerted on a car executing a turn on a banked track is given by the expression:
[tex]N=\frac{mg}{cos \theta - \mu sin \theta}[/tex]
where m is the mass of the car, g is the gravitational acceleration, [tex]\theta[/tex] is the angle of the bank, and [tex]\mu[/tex] is the coefficient of friction.
From the formula, we see that the normal force depends on [tex]\theta[/tex] (the angle of the bank) and [tex]\mu[/tex] (the coefficient of friction), while it does not depend on the speed of the car or on the radius of curvature. Therefore, these two are the correct answers.
