A disc spins at a rate of 2 rps. There is a 15 g object on the disc located 10 cm away from the center. Find the object's tangential velocity and the frictional force that has to be exerted on the object to keep it from slipping off the disc.
My answer:
Tangential velocity = 10*2 = 20 cm/s
Frictional force = centripetal force = mv^2/r = 600 Newtons
If 'rps' here stands for revolutions per second, then you must convert it to radians per second first, otherwise 20cm/sec is correct
Centripetal force = mv^2/r
= 0.015*0.2*0.2/0.1
= 0.015*0.4
= 0.006 N
Gotta convert your units
Circumference = pi*2r = 3.14 * 20 = 62.8 cm.
V = 62.8m/rev. * 2rev/s = 125.7 m/s.
Correction: V = 62.8cm/rev * 2rev/s = 125.7 cm/s.
To find the object's tangential velocity, you need to multiply the given rotational rate of 2 revolutions per second (rps) by the object's distance from the center. In this case, the object is located 10 cm away from the center. Therefore, the tangential velocity can be calculated by multiplying 2 rps by 10cm:
Tangential velocity = 2 rps * 10 cm = 20 cm/s.
To find the frictional force required to keep the object from slipping off the disc, you need to consider the centripetal force acting on the object. The centripetal force is given by the formula mv^2/r, where m is the mass of the object, v is its tangential velocity, and r is its distance from the center.
The mass of the object is given as 15 g. To convert this to kilograms, divide by 1000:
m = 15 g / 1000 = 0.015 kg.
Substituting the values into the centripetal force formula:
Centripetal force = (0.015 kg) * (20 cm/s)^2 / 10 cm.
Simplifying the equation:
Centripetal force = (0.015 kg) * (400 cm^2/s^2) / 10 cm.
Now, let's convert cm^2/s^2 to m^2/s^2 by dividing by 100^2:
Centripetal force = (0.015 kg) * (4 m^2/s^2) / 10 cm.
Centripetal force = (0.015 kg) * (4 m^2/s^2) / 0.1 m.
Finally, multiplying the values:
Centripetal force = 0.06 kg * m^2/s^2 / 0.1 m = 0.6 kg * m/s^2.
Therefore, the frictional force required to keep the object from slipping off the disc is 0.6 Newtons.