(1) A block of mass 1.76kg is on a wedge with a block-wedge static coefficient o
ID: 2236924 • Letter: #
Question
(1) A block of mass 1.76kg is on a wedge with a block-wedge static coefficient of friction, ?s= 0.184. The wedge angle ? = 25.5ois as shown in the above illustration. The wedge is rotated with an angular frequency of ? with the block's center of mass located a distance L = 0.452m up the ramp from the axis of rotation as depicted by the above diagram. (a) What is the largest positive value of ? that is allowed to keep the block at the same location on the wedge? (b) What is the smallest positive value of ? that is allowed to keep the block at the same location on the wedge?
(2) A penny of mass 3.11g is placed face-side-up on a record that is on a record player. The coefficient of static friction is ?s= 0.314. (a) If the penny is initially a distance r = 2.15cm from the axis of rotation, then at what angular frequency, ?, does the penny begin to move? (b) If the penny that is still placed face-side-up is on the same record, but now rotating at an angular velocity of 7.1rad/s, then what is the maximum distance that the penny's center of mass can be from the axis of rotation without moving? Let us now flip the penny to face-side-down, where the new coefficient of friction is 0.454. (c) If the penny sitting face-side-down is initially a distance r = 2.83cm from the axis of rotation, then at what angular frequency, ?, does the penny now begin to move? (d) If the penny that is still placed face-side-down is on the same record, but now rotating at an angular velocity of 7.2rad/s, then what is the maximum distance that the penny's center of mass can be from the axis of rotation without moving?
(3) A spherically symmetric object with a density of 1.37g/cm3and a radius of 13.5cm is falling. Assuming is has a drag coefficient, D, of 0.368, (a) what is the magnitude of terminal velocity? (b) If we were to now ignore the air's resistance to the falling object, then at what height must the object be dropped to achieve this value?
Explanation / Answer
1,,,,change the values a. Use conservation of momentum. So, write out the momentum of the block and set it equal to the momentum of the wedge. mBvB=mWvW (.5)(4)=(3)(vW) vW=.667 m/s vW= -.667 m/s The wedge is going to go the opposite direction as the block, so make it negative. b. Use conservation of energy. The gravitational potential energy equals the kinetic energy of the block. mgh=1/2mvB^2 9.8h=1/2(4)^2 h=.816 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.