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Significance

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case.

Check Your Understanding The fan blades on a jet engine have a moment of inertia 30.0 kg-m 2 . In 10 s, they rotate counterclockwise from rest up to a rotation rate of 20 rev/s. (a) What torque must be applied to the blades to achieve this angular acceleration? (b) What is the torque required to bring the fan blades rotating at 20 rev/s to a rest in 20 s?

a. The angular acceleration is α = 20.0 ( 2 π ) rad / s 0 10.0 s = 12.56 rad / s 2 . Solving for the torque, we have i τ i = I α = ( 30.0 kg · m 2 ) ( 12.56 rad / s 2 ) = 376.80 N · m ; b. The angular acceleration is α = 0 20.0 ( 2 π ) rad / s 20.0 s = −6.28 rad / s 2 . Solving for the torque, we have i τ i = I α = ( 30.0 kg-m 2 ) ( −6.28 rad / s 2 ) = −188.50 N · m

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Summary

  • Newton’s second law for rotation, i τ i = I α , says that the sum of the torques on a rotating system about a fixed axis equals the product of the moment of inertia and the angular acceleration. This is the rotational analog to Newton’s second law of linear motion.
  • In the vector form of Newton’s second law for rotation, the torque vector τ is in the same direction as the angular acceleration α . If the angular acceleration of a rotating system is positive, the torque on the system is also positive, and if the angular acceleration is negative, the torque is negative.

Conceptual questions

If you were to stop a spinning wheel with a constant force, where on the wheel would you apply the force to produce the maximum negative acceleration?

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A rod is pivoted about one end. Two forces F and F are applied to it. Under what circumstances will the rod not rotate?

If the forces are along the axis of rotation, or if they have the same lever arm and are applied at a point on the rod.

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Problems

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?

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Suppose you exert a force of 180 N tangential to a 0.280-m-radius, 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

a. τ = ( 0.280 m ) ( 180.0 N ) = 50.4 N · m ; b. α = 17.14 rad / s 2 ;
c. α = 17.04 rad / s 2

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A flywheel ( I = 50 kg-m 2 ) starting from rest acquires an angular velocity of 200.0 rad/s while subject to a constant torque from a motor for 5 s. (a) What is the angular acceleration of the flywheel? (b) What is the magnitude of the torque?

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A constant torque is applied to a rigid body whose moment of inertia is 4.0 kg-m 2 around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.0 s, what is the applied torque?

τ = 8.0 N · m

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A torque of 50.0 N-m is applied to a grinding wheel ( I = 20.0 kg-m 2 ) for 20 s. (a) If it starts from rest, what is the angular velocity of the grinding wheel after the torque is removed? (b) Through what angle does the wheel move through while the torque is applied?

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A flywheel ( I = 100.0 kg-m 2 ) rotating at 500.0 rev/min is brought to rest by friction in 2.0 min. What is the frictional torque on the flywheel?

τ = −43.6 N · m

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A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. (a) What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions? (b) A tool whose coefficient of kinetic friction with the wheel is 0.60 is pressed perpendicularly against the wheel with a force of 40.0 N. What torque must be supplied by the motor to keep the wheel rotating at a constant angular velocity?

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Suppose when Earth was created, it was not rotating. However, after the application of a uniform torque after 6 days, it was rotating at 1 rev/day. (a) What was the angular acceleration during the 6 days? (b) What torque was applied to Earth during this period? (c) What force tangent to Earth at its equator would produce this torque?

a. α = 1.4 × 10 −10 rad / s 2 ;
b. τ = 1.36 × 10 28 N-m ; c. F = 2.1 × 10 21 N

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A pulley of moment of inertia 2.0 kg-m 2 is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. What are (a) the angular acceleration of the pulley and (b) the linear acceleration of the weights? Assume the following data: r 1 = 50 cm , r 2 = 20 cm , m 1 = 1.0 kg , m 2 = 2.0 kg .

Figure shows a pulley mounted on a wall. Light strings are wrapped around two circumferences of the pulley and weights are attached. Smaller weight m1 is attached to the outer circumference of radius r1. Larger weight M2 is attached to the inner circumference of radius r2.
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A block of mass 3 kg slides down an inclined plane at an angle of 45 ° with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). The pulley can be approximated as a disk. The coefficient of kinetic friction on the plane is 0.4. What is the acceleration of the block?

Figure shows a block that slides down an inclined plane at an angle of 45 degrees with a tether attached to a pulley.

a = 3.6 m / s 2

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The cart shown below moves across the table top as the block falls. What is the acceleration of the cart? Neglect friction and assume the following data: m 1 = 2.0 kg , m 2 = 4.0 kg , I = 0.4 kg-m 2 , r = 20 cm

Figure shows the pulley installed on a table. A cart of mass m2 is attached to one side of the pulley. A weight m1 is attached at another side and hangs in air.
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A uniform rod of mass and length is held vertically by two strings of negligible mass, as shown below. (a) Immediately after the string is cut, what is the linear acceleration of the free end of the stick? (b) Of the middle of the stick?

Figure shows a rod that is held vertically by two strings connected at its ends. One of the strings is cut with scissors.

a. a = r α = 14.7 m / s 2 ; b. a = L 2 α = 3 4 g

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A thin stick of mass 0.2 kg and length L = 0.5 m is attached to the rim of a metal disk of mass M = 2.0 kg and radius R = 0.3 m . The stick is free to rotate around a horizontal axis through its other end (see the following figure). (a) If the combination is released with the stick horizontal, what is the speed of the center of the disk when the stick is vertical? (b) What is the acceleration of the center of the disk at the instant the stick is released? (c) At the instant the stick passes through the vertical?

Figure A shows a thin stick attached to the rim of a metal disk. Figure B shows a thin stick that is attached to the rim of a metal disk and rotates around a horizontal axis through its other end.
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Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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