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  • Topic: Torque, Moment of inertia, Angular momentum
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  • Published : November 10, 2012
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SAMPLE PROBLEMS: 111-SET #9 ROTATIONAL MOTION PROBLEMS: 09-1

1) A grinding wheel starts from rest and has a constant angular acceleration of 5 rad/sec2. At t = 6 seconds find the centripetal and tangential accelerations of a point 75 mm from the axis. Determine the angular speed at 6 seconds, and the angle the wheel has turned through.

|We have a problem of constant angular acceleration. The figure & coordinate system are |[pic] | |shown. Since a time is given in the problem we must use the equations of motion. | | | | | |((t) = (1/2) ( t2 + ( 0 t ; ((t) = ( t + ( 0 | | | | | |The initial state of motion is: ( 0 = 0; ( 0 = 0 . We are given ( in the problem. | | |Hence we have the specific equations of motion: | |

((t) = (1/2)(5) t2 ; ((t) = 5 t .

Thus at time t = 6 sec, ((6 sec) = (5)(6) = 30 rad/sec . And: ((6 sec) = (1/2)(5)(6)2 = 90 rad .

The centripetal & tangential accelerations are linear quantities. Hence, to calculate we need the interconnecting equations. ( s = r (( ; v = r ( ; a t = r (

(Where all angular quantities are expressed in terms of angular units of radians.) Thus, at 6 seconds we have: for velocities for acceleration

v(6 sec) = r ( = (.075)(30) = 2.25 m/sec a t = r ( = (.075)(5) = .375 m/sec2

a c = v2/r = r ( 2 = (.075)(30)2 = 67.5 m/sec2 .

The magnitude of the total linear acceleration of this point would be given by:

[pic]

2) An electric motor operates at 1800 rpm. Find its angular speed in radians/second. What is the linear speed of a point 55 mm from the axis of rotation? What is its centripetal acceleration?

Solution: We have a rotational problem of constant angular velocity (UCM). The quantity 1800 rpm (rev/minute) can be interpreted as a frequency, or as an angular velocity. The equation: ( = 2 ( f converts 'frequency' into an angular velocity in units of radians/unit time. Thus:

09-2
( = 2 ( (1800) rad/min = 188 rad/sec .

The linear speed of a point is related to the angular velocity through the inter-connecting equation:

v = r ( = (.055)(188) = 10.34 m/sec; a t = r ( = 0 (since ( is constant)

ac = v2/r = r ( 2 = (.055)(188)2 = 1943.9 m/sec2

3) The turbine and associated rotating parts of a jet engine have a total moment of inertia of 25 kg-m2. It is accelerated uniformly from rest to an angular speed of 150 rad/sec in a time of 25 seconds. Find (a) the angular acceleration; (b) the net torque required; (c) the angle turned through in 25 seconds; (d) the work done by the net torque; and (e) the kinetic energy of the turbine at the end of the 25 seconds.

|Solution: Since time is explicitly mentioned in the problem we expect that we will have to |[pic] | |employ the equations of motion. Setting up the CS as shown, & with the initial conditions: | | | |...
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