Answer:
Explanation:
The torque of a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation of the system
is the angle between the direction of the force and d
In this problem, we have:
, the force
, the distance of application of the force from the centre (0,0)
, the angle between the direction of the force and a
Therefore, the torque is
Answer:
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Explanation:
Given data,
The mass of the bar AB, m = 6.4 kg
The angular velocity of the bar, θ˙ = 2.7 rad/s
The angle of the bar at A, θ = 24°
Let the length of the bar be, L = l
The angular moment at point A is,
∑ Mₐ = Iα
Where, Mₐ - the moment about A
α - angular acceleration
I - moment of inertia of the rod AB
Let G be the center of gravity of the bar AB
The position vector at A with respect to the origin at G is,
The acceleration at the center of the bar
Since the point A is fixed, acceleration is 0
The acceleration with respect to the coordinate axes is,
Comparing the coefficients of i
Comparing coefficients of j
Net force on x direction
substituting the values
=1.5(14.58L+11.96)
Similarly net force on y direction
= 3.2(2.97L - 157.03) + 62.72
Where L is the length of the bar AB
Therefore the net force,
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Substituting the value of L gives the force at pin A
Answer:
the ball experiences the greater momentum change
Explanation:
You have to take into account that momentum change is given by
where vf and vb are the speed of the object after and before the impact.
In the case of the ball you have
where the minus of vf is included due to the motion is in an opposite direction regarding with vb
And for the lump
Hence, the ball experiences the greater change
hope this helps!!