(1) A uniform meter rod 4ON has a piece of weight of 2ON attached to one of its end. What is the center of gravity of the system from the weight end? (2) two bodies X and Y start from rest and move with uniform acceleration of a and 4a respectively. If the bodies cover the same distance in times T(x) and T(y) find the ratio of T(x) and T(y)?
a. vertical forces:
fu=40+20
now, summing moments about the weighted end:
40*.5L-60*xL=0
x=20/60 so the cb is 1/3 the length from the weight end.
b. Same distance,
d1=1/2 a Tx^2
d2=1/2 4a Ty^2
same distance, so
(Tx/Ty)^2 =4
Tx/Ty=2
(1) To find the center of gravity of the system, we need to consider the distribution of mass along the meter rod.
The center of gravity, also known as the center of mass, is the point where the entire mass of an object or system can be considered to be concentrated. In this case, we have a uniform meter rod with a weight attached to one end.
Given:
- Length of the meter rod: 4 m
- Weight attached: 2 ON
Since the meter rod is uniform, the mass is distributed evenly along its length. The 4 m rod has a total weight of 4 ON.
To find the center of gravity, we need to consider the weight of each segment of the rod.
Let's divide the meter rod into two segments: segment AB (with weight 2 ON) and segment BC (with weight 2 ON). The weight of each segment is distributed evenly along its length.
The center of gravity can be calculated using the formula:
Center of gravity = Sum of (mass * position) / Total mass
For segment AB, the center of gravity is at the midpoint:
Center of gravity of AB = (2 ON * 1 m) / 2 ON = 1 m
For segment BC, the center of gravity is at the midpoint as well:
Center of gravity of BC = (2 ON * 3 m) / 2 ON = 3 m
Since the center of gravity of the system is the weighted average of the center of gravity of each segment, we can calculate it as follows:
Center of gravity of the system = (Weight(AB) * Center of gravity(AB) + Weight(BC) * Center of gravity(BC)) / Total weight
Center of gravity of the system = (2 ON * 1 m + 2 ON * 3 m) / 4 ON = 2 m
Therefore, the center of gravity of the system from the weight end is 2 m.
(2) Given:
Acceleration of body X: a
Acceleration of body Y: 4a
Let's assume that both bodies cover the same distance, which we can represent as d.
Using the second equation of motion:
d = ut + (1/2)at^2
Since both bodies start from rest, the initial velocity (u) is zero.
For body X:
d = (1/2)a(T(x))^2 ----(1)
For body Y:
d = (1/2)(4a)(T(y))^2 ----(2)
Divide equation (1) by equation (2):
(T(x))^2 / (T(y))^2 = (a/4a)
Simplifying:
(T(x))^2 / (T(y))^2 = 1/4
Taking the square root:
T(x) / T(y) = 1/2
Therefore, the ratio of T(x) to T(y) is 1:2.