What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? the axis of rotation is perpendicular to a clock face and through the center of the clock. (b) what is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 p. m.?

a) 0 Nm b) mgL/2 Nm counterclockwise

Explanation:

Before we even begin answering the question, we have to appreciate the fact that the weight of the constant density symmetrical rigid objects is considered to be acting from its geometrical center, which we will call the center of mass. If the clock arm is considered to be rectangular , rigid and constant density(assumed, since this is a high school question) we can proceed with our analysis. We also note that the torque on the arm are given by,

where m is the total mass of the arm , x is the horizontal distance(perpendicular to the direction of force) of the center of mass of the arm from the center of the clock and g is the acceleration due to gravity. Also let L be the length of the hour arm.

In part a) the horizontal distance of the center of mass of the arm from the center of the clock would be zero since at noon the center of mass of the clock arm is right above the center of the clock. So x = 0 , hence T = 0 Nm.

in part b) the horizontal distance between the center of mass of the arm and the center of the clock is L/2 , where L is the length of the hour arm , but the weight acts from half that distance since the weight of the object is considered to be acting from the geometrical center of the object which in this case is L/2 distance away from the center of the clock. So plugging into the above equation we get T = mg*(L/2) Nm, the force acting downwards from the left since the arm is pointing at 9:00 , which will result in a counter clockwise torque. So T=mgL/2 Nm counter clockwise is the Torque due to the hour arm.

If you already know calculus you can check the answer to part b) by evaluating the integral,

where M is the mass per unit length of the arm. m is the total mass given by m=ML.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.