Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let a = 200 N and b = 400 N. Round y
our answers to one decimal place.)
1 answer:
Answer: R = 346.4N and angle 30° to the horizontal negative axis
Explanation:
To find the resultant force, we need to sum up the forces on the vertical and horizontal axis.
For the horizontal axis;
Rx = -b + acos60
Rx = -400N +200cos60
Rx = -400N +100N
Rx = -300N
For the vertical axis;
Ry = asin60 = 200sin60
Ry = 173.2N
The resultant force R can be given as;
R = √(Rx^2 +Ry^2)
R = √((-300)^2 + 173.2^2)
R = 346.4N
Angle z can be written as
Tanz = Ry/Rx
z = taninverse (Ry/Rx)
z = taninverse (173.2/300)
z = 30°
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Answer:
41.667 per cent of the total kinetic energy is translational kinetic energy.
Explanation:
As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:
Where:
- Total energy, measured in joules.
- Rotational kinetic energy, measured in joules.
- Translational kinetic energy, measured in joules.
The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:
Rotational kinetic energy
Translational kinetic energy
Where:
- Moment of inertia of the spherical shell with respect to its center of mass, measured in
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- Angular speed of the spherical shell, measured in radians per second.
- Radius of the spherical shell, measured in meters.
After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:
In addition, the moment of inertia of a spherical shell is equal to:
Then, total energy is reduced to this expression:
The fraction of the total kinetic energy that is translational in percentage is given by the following expression:
41.667 per cent of the total kinetic energy is translational kinetic energy.