R = 1.1 metersM= 286 kgm = 41 kgv = 2.5 m/sPart (a) Calculate the moment of inertia of the merry-go-round, in kg m²./ Correct!I= 178Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of themerry-go-round./ Correct!01 = 2.27Part (c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.w2 = 0.013134w2 = 0.01313Part (d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round?w3 = 0.0158回於is at the center of the merry go round?Part (e) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boyw4 = 1.004 = 1Part (f) Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow,he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of themerry-go-round after the boy jumps off?ws = 0.015|sin()cos()tan()89HOMEcotan()asin()acos()E4atan()acotan()sinh()123cosh()tanh()cotanh()ENDO Degrees O RadiansBACKSPACEDELCLEARSubmitHintFeedbackI give up!

Since the question contains more than three subparts, solution for the first three is provided here.…

R= 1.1 meters M= 286 kg m = 41 kg v = 2.5 m/s Part (a) Calculate the moment of inertia of the merry-go-round, in kg · m2. / Correct! I- 178 Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. j-מ!2.274Correct Part (c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. w2 = 0.013134 : 13 0.013 - נ" Part (d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry- go-round when the boy is half way between the edge and the center of the merry go round? w3 = 0.0158 Part (e) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? w4 = 1.0 Part (f) Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off? ws = 0.015| sin() cos() tan() 8 9 HOME cotan() asin() acos() E 1^ 4 5 6 atan acotano sinh0 1 3

R= 1.1 meters M= 286 kg m = 41 kg v = 2.5 m/s Part (a) Calculate the moment of inertia of the merry-go-round, in kg · m2. / Correct! I- 178 Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. j-מ!2.274Correct Part (c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. w2 = 0.013134 : 13 0.013 - נ" Part (d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry- go-round when the boy is half way between the edge and the center of the merry go round? w3 = 0.0158 Part (e) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? w4 = 1.0 Part (f) Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off? ws = 0.015| sin() cos() tan() 8 9 HOME cotan() asin() acos() E 1^ 4 5 6 atan acotano sinh0 1 3

Similar questions

Fe -4i+4j+2k LB 6IN Determine the moment due to the force about point 0, expressing the moment as a Cartesian Vector.
A rotating wheel requires 2.93-s to rotate through 37.0 revolutions. Its angular speed at the end of the 2.93-s interval is 98.5 rad/s. What is the constant angular acceleration of the wheel? | rad/s²
A figure skater performs a twisting jump. She rotates around her longitudinal axis 2 times in 0.65 s. What was her average angular velocity in radians/second? A 720rad/s B 19.3rad/s C 4? rad/s D 9.67 rad/s E 1107 rad/s
A figure skater performs a twisting jump. She rotates around her longitudinal axis 2 times in 0.65 s. What was her angular displacement in radians? A 4 В D 720 E 2m
The uniform 22-kg plate is released from rest at the position shown. (Figure 1) Figure A x 0.5 m- 1 of 1 0.5 m > Part A Determine its initial angular acceleration. Express your answer in radians per second squared to three significant figures. Enter positive value if the angular acceleration is clockwise and negative value if the angular acceleration is counterclockwise. a = Submit Part B 195| ΑΣΦ. 41 Request Answer vec ? rad/s² Review Determine the x and y components of reaction at the pin A using scalar notation. Express your answers in newtons to three significant figures separated by a comma.
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. v= 35.5 m/s d=0.29 m If the velocity of the ball in the pitcher's hand is 35.5m/s and the ball is 0.29 m from the elbow joint, what is the angular velocity (rad/s) of the foerarm?
The force of F = (2i -5j) N is acting on a particle whose position vector is r = (i +0.5j +2k) m Calculate the torque vector about the origin in Nm (Hint: consider the vector product) a. 8 b. 10i +4j - 6k c. 2i-2.5J d. (-1k) e. O f. (k) Nm 2i -2.5j +2k
The angular position of a rotating object as a function of time is given below: e = 3.113-2.5t2 +4.7t+d Here e is in units of radians and t is in units of seconds and d is constant. Determine the instantaneous angular acceleration of the object at t=2.9s. Express your answer in units of rad/s using one decimal place.
A 3.00-kg mass is placed at (2.00, 3.00) m, and a 4.50-kg mass is placed at (2.00, -3.00) m. What is the moment of inertia of this system of masses about the x-axis? (Take the rotational axis to be on the x-axis) 15 kg m² O 67.5 kg*m² O 30 kg*m² O 0kg*m² O97.5 kg*m²
4. A skater has a moment of inertia of 105.0 kg.m² when his arms are outstretched and a moment of inertia of 64.0 kg.m² when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 80.0 rpm (revolutions per minute) with his arms outstretched, what will his angular speed be when they are tucked in? rpm
given:wheel, radius 6.6 cm, m = 4.2 kgaccelerates from rest with angular acceleration 2.6 rad/s2.After 27 s, what is the angular velocity?
The massive pully has moment of inertia I = 1/2MR2. Calculate the downward acceleration of the hanging mass (as a positive number, in m/s²). %3D • M = 700 grams %3D • m = 70 grams • R = 9 cm M, R m