PHSX 220 Homework 10 Paper Due Wed. April 6 5:00 pm Rotation, Torque, Energy and

Question

PHSX 220 Homework 10 Paper Due Wed. April 6 5:00 pm Rotation, Torque, Energy and Angular Momentum Problem!: Shown below is a solid disc of mass 100 g (M = 100 g), and radius of 10 cm (1-10 cm) and moment of incrtia about its center of mass cqual to Lo 1/2MR. The disc is initially rotating at 10 revolutions per second into the page as shown in the top view of the figure. A constant force of 10 N is being applied always tangent to the disc a distance of 2R away from the axis of rotation. Answer the following questions in regards to the configuration. Do not consider any friction or forces due to gravity in this problem. The top view will be the reference for all dircctions in this problem rotational variable directions will be into the page or out of the pago. a) Calculate the magnitude and direction of the initial angular velocity b) Calculate the magnitude and direction of the torque acting on the disc c) Caleulate the moment of inertia about the rotation axis d) Caleulate the angular acceleration of the disc e) Caleulate the magnitude and direction of the dises angular velocity after 1 second f) Calenlate the rotational kinetic energy of the dise after 1 second g) Calculate the angular momentum of the disc after 1 second 11) Graph ,w, vs. time Pirot (into and out of page) a Top View Side view

Explanation / Answer

(a)

angular velocity w = 10*2*pi rad/s on = 62.83 rad/s

direction clockwise

(b)

torque = 2R*F = 2*0.1*10 = 2 Nm

direction counter clockwise

(c)

I = (1/2)*M*R^2 + M*R^2

I = (3/2)*M*R^2 = (3/2)*0.1*0.1^2 = 0.0015 kg m^2

(d)

torque = I*alpha

2 = 0.0015*alpha

alpha = 1333.33 rad/s^2

(e)

alpha = (w2-w1)/t

1333.33 = (w2-62.83)/1

w2 = 1396 rad/s

f)

KE = 0.5*I*w2^2

KE = 0.5*0.0015*1396^2

KE = 1461.612 J

g)L = I*w2 = 0.0015*1396 = 2.094 kg m^2 /s

3)

initial potential energy of the system Ui = 2M*g*d

initial kinetic energy Ki = 0

final potential energy Uf = 0

final KE , Kf = 0.5*I*w^2 + 0.5*I*w^2 + 0.5*2M*v^2

Kf = 0.5*0.5*M*R^2*w^2 + 0.5*0.5*4M*(2R)^2 + 0.5*2M*v^2

v = Rw

Kf = 0.25*M*v^2 + 0.25*4*4*M*v^2 + 0.5*2M*v^2

Kf = M*v^2 (0.25 + 4 + 1)

Kf = 5.25*M*v^2

from energy conservation

2M*g*d = 5.25*M*v^2

v = 0.4*g*d <<----answer

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