Two masses (mA= 2 kg, mB= 4 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.
c.) Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level*?
? N
d.) Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant*.
? rad/s2
There has to be a center of rotation it seems to me, and it is not specified.
I think the assumption is to assume that the stick is level with masses A and B attached.
Thus if the balance point is x cm from the zero
2x=4(75-x) so x=50 cm
Assuming this then to keep it level
2 kg are needed at 100 cm.
If g=10 m s^-2 then the force is 20 N.
Not a very well worded question in my view.
To find the force needed to keep the meter stick level when mass B is removed, we need to consider the torque balance.
c.) The torque balance equation is given by:
Sum of torques = 0
Since the meter stick is level, the sum of torques about any point is equal to zero. Let's choose the 0 cm mark as the reference point.
The torque due to mass A about the reference point is given by:
Torque_A = (distance to reference point)*(mass of A)*(acceleration due to gravity)
= (0.75 m)*(2 kg)*(9.8 m/s^2)
= 14.7 N·m
Since there is no mass at the 100 cm mark, the torque due to the force exerted at the 100 cm mark is:
Torque_force = (distance to reference point)*(force)
= (1.0 m)*(force)
The sum of torques is zero, so:
Torque_A + Torque_force = 0
14.7 N·m + (1.0 m)*(force) = 0
Solving for the force:
force = -14.7 N·m / 1.0 m
force ≈ -14.7 N
Therefore, a force of approximately 14.7 N in the opposite direction of mass A needs to be exerted at the 100 cm mark to keep the meter stick level when mass B is removed.
d.) To calculate the angular acceleration of the meter stick when mass B is removed and no additional force is supplied, we can use the torque and moment of inertia.
The moment of inertia of a meter stick about its center of mass (assumed to be in the middle) is given by:
I = (1/12)*(mass of meter stick)*(length of meter stick)^2
The mass of the meter stick is not provided, so the value of I cannot be determined. Without the value of I, we cannot calculate the angular acceleration.
To find the force required at the 100 cm mark to keep the meter stick level when mass B is removed (part c), we need to consider the torque created by the remaining mass A.
First, we need to find the torque created by mass A. Torque is the rotational force that causes an object to rotate. It is calculated as the product of the force applied at a distance from the pivot point.
In this case, the force applied by mass A is its weight, which is given by the formula F = m * g, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The distance between the pivot point (0 cm mark) and the force applied by mass A (0 cm mark) is zero. Therefore, the torque created by mass A is zero.
Now, let's move on to part d, where we need to calculate the angular acceleration of the meter stick when mass B is removed and no additional force is supplied.
In order to calculate the angular acceleration, we need to use the equation τ = I * α, where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
The moment of inertia of a meter stick rotating about one end is given by the formula I = (1/3) * m * L^2, where m is the mass of the meter stick and L is the length of the meter stick.
In this case, the mass of the meter stick (mA) is 2 kg, and the length (L) is 100 cm (or 1 meter).
Substituting the values into the equation, we get:
I = (1/3) * 2 kg * (1 m)^2 = 2/3 kg.m^2
Since no external torque is applied (as mentioned in the question), the total torque is zero. Therefore, we can set the equation τ = I * α to zero:
0 = (2/3) kg.m^2 * α
Solving for α, we get:
α = 0 rad/s^2
Hence, the angular acceleration of the meter stick, when mass B is removed and no additional force is supplied, is 0 rad/s^2.