1.A 35-kg child is sitting along the rim of a merry-goround that is rotating at 0.35 revolutions per second about its symmetry axis. The mass and the radius of the merry-go-round are 85 kg and 8 m, respectively. Assume that you can treat the child as a point particle and you can model the merry-go-round as a disc. a. Calculate the moment of inertia of the system about its axis of symmetry b. Calculate the total angular momentum of the system. A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall. If the ladder is in a state of impending motion when it makes an angle of 65° with respect to the floor, calculate the force exerted by the wall onto the ladder. Solution and Answer: Since the ladder is uniform, then the center of gravity of the ladder is at its geometric center. We use the first and second conditions for equilibrium in order to tackle this problem. The chosen axis of rotation is at the point where the ladder makes contact with the ground. Let ????? and ? be the forces due to the wall and due to friction, respectively. From the first conditions of equilibrium, we get: ?????) = 0 The second condition for equilibrium is: ????? + (−?????????) + ?? + (−?????ℎ?) = 0 The frictional force and the normal force are located at the pivot point. This further simplifies to: ????? + (−?????ℎ?) = 0 (?????)(5 m)(sin65°) = (?)(2.5 m)(sin25°) Now, compute for the value of the force exerted by the wall on the ladder,
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
1.A 35-kg child is sitting along the rim of a merry-goround that is rotating at 0.35 revolutions per second about its symmetry axis. The mass and the radius of the merry-go-round are 85 kg and 8 m, respectively. Assume that you can treat the child as a point particle and you can model the merry-go-round as a disc.
a. Calculate the moment of inertia of the system about its axis of symmetry
b. Calculate the total
A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall. If the ladder is in a state of impending motion when it makes an angle of 65° with respect to the floor, calculate the force exerted by the wall onto the ladder. Solution and Answer:
Since the ladder is uniform, then the center of gravity of the ladder is at its geometric center. We use the first and second conditions for equilibrium in order to tackle this problem. The chosen axis of rotation is at the point where the ladder makes contact with the ground. Let ????? and ? be the forces due to the wall and due to friction, respectively. From the first conditions of equilibrium, we get:
?????) = 0
The second condition for equilibrium is: ????? + (−?????????) + ?? + (−?????ℎ?) = 0
The frictional force and the normal force are located at the pivot point. This further simplifies to:
????? + (−?????ℎ?) = 0
(?????)(5 m)(sin65°) = (?)(2.5 m)(sin25°)
Now, compute for the value of the force exerted by the wall on the ladder,
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