Question
1. A 0.20 kg object moves at a constant speed in a horizontal circular path of radius "r" while suspended from the top of a pole by a string of length 1.2 m. It makes an angle of 62 degrees with the horizontal. What is the speed of this object?
2. An object of mass 3 kg is traveling in a horizontal circular path of radius 1.2 m while suspended by a piece of string length 1.9 m. It makes an angle of 39 degrees with the horizontal. What is the centripedal force on the mass?
2. An object of mass 3 kg is traveling in a horizontal circular path of radius 1.2 m while suspended by a piece of string length 1.9 m. It makes an angle of 39 degrees with the horizontal. What is the centripedal force on the mass?
Answers
The object is in equilibrium due to three forces: Tension T of the string, centripetal force C, and weight W.
Since C and W are orthogonal (at right angles), T can be resolved into two components in these directions, each opposing C and W.
In brief, tan(62°)=W/C
Thus the vertical component of T is mg.
The horizontal component is the centripetal force, mv²/r, where r is the horizontal radius, r=Lcos(θ), where L=length of string, and θ=62°.
Thus
mg/(mv^2/(1.2*cos(62°)))=tan(62°)
Solve for v. I get v=1.714 m/s
Since C and W are orthogonal (at right angles), T can be resolved into two components in these directions, each opposing C and W.
In brief, tan(62°)=W/C
Thus the vertical component of T is mg.
The horizontal component is the centripetal force, mv²/r, where r is the horizontal radius, r=Lcos(θ), where L=length of string, and θ=62°.
Thus
mg/(mv^2/(1.2*cos(62°)))=tan(62°)
Solve for v. I get v=1.714 m/s
when an object moves in a circular path it changes direction and is therefore accelerating a force pushes the object toward the center of the circle what is the name of that force