12 1 Model A spinning skater whose arms are outstretched is a rigid rotating body Visualize v where Solve The speed v 0 70 m 6 rad s 13 2 m s Assess A speed of 13 2 m s 26 mph r r 140 cm 2 0 70 m Also 180 rpm 180 2 60 rad s 6 rad s Thus for the hands is a little high but reasonable 12 2 Model Assume constant angular acceleration Solve a The final angular velocity is f 2000 rpm 209 4 rad s The definition of 2 rad min 60 s rev angular acceleration gives us f i t t 209 4 rad s 0 rad s 0 50 s 419 rad s The angular acceleration of the drill is b 2 4 2 10 rad s 1 2 rev 2 rad The drill makes 52 4 rad 8 3 revolutions i i t f t 2 0 rad 0 rad 419 rad s 0 50 s 2 52 4 rad 1 2 12 3 Model Assume constant angular acceleration Visualize Solve a Since r find first With 90 rpm 9 43 rad s and 60 rpm 6 28 rad s ta t 9 43 rad 6 28 rad s 10 s 0 314 rad s 2 The angular acceleration of the sprocket and pedal are the same So 2 0 18 m 0 314 rad s r ta 0 057 m s 2 b The length of chain that passes over the sprocket during this time is L r Find i i t f 1 2 t 2 i f 6 28 rad s 10 s 0 314 rad s 2 10 s 2 78 5 rad 1 2 The length of chain which has passed over the top of the sprocket is L 0 10 m 78 5 rad 7 9 m 12 4 Model Assume constant angular acceleration Visualize Solve The initial angular velocity is i 60 rpm The angular acceleration is 2 rad min 60 s rev 2 rad s f i t 0 rad s 2 rad s 25 s 0 251 rad s 2 The angular velocity of the fan blade after 10 s is i f t t 0 2 rad s 0 251 rad s 2 10 s 0 s 3 77 rad s The tangential speed of the tip of the fan blade is tv r 0 40 m 3 77 rad s 1 51 m s b i i t f t 2 0 rad 2 rad s 25 s 0 251 rad s 2 25 s 2 78 6 rad 1 2 1 2 The fan turns 78 6 radians 12 5 revolutions while coming to a stop 12 5 Model The earth and moon are particles Visualize sets the coordinate origin at the center of the earth so that the center of mass location is the 0 m Ex Choosing distance from the center of the earth Solve x cm M M m x m x E E m m E 4 67 10 m M 6 24 5 98 10 kg 0 m 7 36 10 kg 3 84 10 m 5 98 10 kg 7 36 10 kg 22 22 24 8 Assess The center of mass of the earth moon system is called the barycenter and is located beneath the surface of the earth Even though E Em is in the denominator of the expression for the earth influences the center of mass location because x cm x 0 m 12 6 Visualize Please refer to Figure EX12 6 The coordinates of the three masses 0 cm 0 cm 0 cm 10 cm and 10 cm 0 cm respectively Solve The coordinates of the center of mass are m m and B A Cm are x cm y cm m x m x C C B B m m C m x A A m A m y m y B B C C m m C m y A A m A B B 100 g 0 cm 200 g 0 cm 100 g 200 g 300 g 300 g 10 cm 5 0 cm 100 g 0 cm 200 g 10 cm 100 g 200 g 300 g 300 g 0 cm 3 3 cm 12 7 Visualize Please refer to Figure EX12 7 The coordinates of the three masses 0 cm 10 cm 10 cm 10 cm and 10 cm 0 cm respectively Solve The coordinates of the center of mass are m m and B A Cm are 200 g 0 cm 100 g 10 cm x cm y cm m x A A m x m x C C B B m m m C A m y m y B B C C m m C B B m y A A m A 300 g 10 cm 200 g 300 g 100 g 300 g 10 cm 200 g 300 g 100 g 200 g 10 cm 100 g 0 cm 6 7 cm 8 3 cm 12 8 Model The balls are particles located at the ball s respective centers Visualize Solve The center of mass of the two balls measured from the left hand ball is x cm 100 g 0 cm 200 g 30 cm 100 g 200 g 20 cm The linear speed of the 100 g ball is v 1 r x cm 0 20 m 120 rev min 2 rad min 60 s rev 2 5 m s 12 9 Model The earth is a rigid spherical rotating body Solve The rotational kinetic energy of the earth is diameter see Table 12 2 is I 2 5 M R earth 2 2 1 I 2 and the angular velocity of the earth is K rot The moment of inertia of a sphere about its Thus the rotational kinetic energy is 2 rad 24 3600 s 5 7 27 10 rad s K rot 2 M R 2 earth 1 2 2 5 1 5 6 5 98 10 kg 6 37 10 m 7 27 10 rad s 5 24 2 2 2 57 10 J 29 12 10 Model The triangle is a rigid body rotating about an axis through the center Visualize Please refer to Figure EX12 10 Each 200 g mass is a distance r away from the axis of rotation where r is given by Solve The moment of inertia of the triangle is frequency of rotation is given as 5 0 revolutions per s or 10 rad s 3 mr I 3 0 200 kg 0 2309 m The rotational kinetic energy is 2 2 0 0320 kg m The 0 20 m r cos30 r 0 2309 m 0 20 m cos30 2 K rot 1 2 1 2 2 I 2 0 0320 kg m 10 0 rad s 2 15 8 J 12 11 Model The disk is a rigid body rotating about an axis through its center Visualize Solve The speed of the point on the rim is given by R The angular velocity of the disk can be v rim determined from its …
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