Mostly Computational 6 A 0.400 kg set of keys is swung in a horizontal circle fr

Question

Mostly Computational 6 A 0.400 kg set of keys is swung in a horizontal circle from the end ofa 0.500 m string around a teacher's head. It takes the keys 1.20 s to go all the way around. ) What is the linear (tangential) speed of the keys? b) What is the tension in the string assuming that the string is perfectly horizontal? If the weight of the keys are considered, the string will slope down from the center of the circle making an angle, , with the horizontal. Why does get smaller as the speed of the keys increases? 7) A constant torque is applied to the rim of a grindstone (disk) whose radius is 0.6 m and mass is 2.5 kg. a) b) What is the grindstone's initial moment of inertia? Find the angular speed after the grind stone has made 12.0 revolutions if. L ii, the magnitude of the torque is 36.0 Nm, and the grindstone starts at rest, then initially speeds up. the torque is 4.00 N·m, and the grindstone starts at 16 rad/s, then initially slows down. c) What is the magnitude of the average friction force applied to the disk initially slowing it down (b-i)? The mass of the Earth is 5.9737076x1024 kg and its average radius is 6,371,912 m. Assume that all of the world's ice is at the poles and does not contribute to the Earth's moment of inertia when in ice form. (Note that some values were changed to make the problem more interesting.) a) Assuming that the Earth is a perfect sphere, what is its moment of inertia? b) If the Earth has a period of 23.93447232 hours, what is its angular velocity? c) If the 2.785 x 10" kg of the world's ice melted and moved to the equator, what would be the the Earth's new moment of 8) inertia? d) What would be the Earth's new angular velocity? e) What is the Earth's new period? 9) A 0.06-cm radius, 3 kg solid ball is rolled from rest down a 20 m inclined plane, which is at an angle of 36.87% Assume no slipping. (a) What is the linear speed of the ball at the bottom of the ramp? (b) What is the angular speed of the ball at the bottom of the ramp?

Explanation / Answer

6)Given,

m = 0.4 kg ; R = 0.5 m ; t = 1.2 s

a)We know that

speed = dist/time

speed = 2 pi R/t = 2 x 3.14 x 0.5/1.2 = 2.617 m/s

Hence, v = 2.617 m/s = 2.62 m/s

b)The tension in horizontal position will be equal to the centripital force

T = Fc = m v^2/R

T = 0.4 x 2.62^2/0.5 = 5.49 N

Hence, T = 5.49 N

c)With the increase in speed the value pf T varies with theta as

T = m v^2/R

Tx = T sin(theta)

Ty = T cos(theta)

theta = tan^-1 (Ty/Tx)

So more the speed less the angle.

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