A mass m1 = 6.0 kg sits on a frictionless table, and rotates at a constant speed
v = 3.0 m/sin a circle of radius R = 1.5 m. m1 is attached by a massless string which
passes through a hole in the centre of the table to a second m2 (which hangs freely
without moving). There is no friction present at the point where the string passes through
the hole in the table. The mass m2 is?
a)1.2kg b)5.2kg c)5.5kg d)2.8kg e)3.7kg
Since there is no motion of the vertical string up or down, the weight of m2 must balance the centripetal force of m1.
m1*V^2/R = m2*g
m2 = m1*V^2/(R*g) = 3.67 kg
That rounds off to answer (e)
To find the mass m2, we can use the concept of centripetal force.
In this scenario, the tension in the string provides the centripetal force required to keep mass m1 moving in a circle. The tension in the string is equal to the weight of mass m2.
First, let's find the centripetal force acting on mass m1 using the formula:
Fc = m1 * v^2 / R
where
Fc = centripetal force
m1 = mass of object 1
v = speed of the object
R = radius of the circle
Plugging in the given values, we have:
Fc = 6.0 kg * (3.0 m/s)^2 / 1.5 m
Solving this equation, we find that the centripetal force Fc is equal to 36.0 N.
Since the tension in the string is equal to the centripetal force, we have:
Tension = Fc = 36.0 N
Now, let's equate the tension in the string to the weight of mass m2:
Tension = weight of m2
Using the formula for weight, we have:
Tension = m2 * g
where
m2 = mass of object 2
g = acceleration due to gravity
Plugging in the given acceleration due to gravity (approximately 9.8 m/s^2) and solving for m2, we get:
36.0 N = m2 * 9.8 m/s^2
m2 = 36.0 N / 9.8 m/s^2
Solving this equation, we find that the mass m2 is approximately 3.7 kg.
Therefore, the answer is e) 3.7 kg.