A clock has a second hand whose tip rubs against the inside of the glass cover. If the frictional force between the glass cover and the tip of the second hand is 0.0020 N and the length of the hand is 9.5 cm, what is the minimum torque that must be applied to the second hand if the clock is not to be stopped?
To determine the minimum torque required to prevent the clock from stopping, we need to consider the effect of static friction between the glass cover and the tip of the second hand.
The torque (T) is given by the formula:
T = F * r
Where:
T is the torque,
F is the force applied, and
r is the distance from the rotation point (pivot) to the point where the force is applied.
In this case, the force applied is the frictional force between the glass cover and the tip of the second hand, which is given as 0.0020 N.
The distance from the rotation point (pivot) to the tip of the second hand is the length of the hand, which is given as 9.5 cm. However, torque is typically measured in meters, so we need to convert the length to meters by dividing it by 100.
r = 9.5 cm / 100 = 0.095 m
Now we can calculate the torque:
T = 0.0020 N * 0.095 m = 0.00019 Nm
Therefore, the minimum torque that must be applied to the second hand to prevent the clock from stopping is 0.00019 Nm.