Physics 1307 Practice Quiz 7 Chapter 10 3.6cm2 1/2
Physics 1307 Practice Quiz 7 Chapter 10 Problem I : Bone Fracture A compressive force of 3.6×10 4 N is exerted on the end of a 20cmlong bone of cross sectional area of
3.6 cm2 , (a) will the bone break, and (b) if not, by how much does it shorten? 6 ( The ultimate compressive strength of bone is about 170×10 9 order of 15×10
N m
2
N m
2
and the Young modules of bone is of the
)
Problem II: Jane to the rescue Tarzan stands on a branch as a leopard threatens. Fortunately, Jane is on a nearby branch of the same hight holding a 25mlong vine attached directly above the point midway between her and Tarzan. How soon does she reach Tarzan. Problem III: Piston motion In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x=5.00cm cos2t where x is in centimeters and t is in seconds. At t=0, (a) find the position of the 6 piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion. SOLUTIONS Problem I To know if the bone is going to break we need to find if the stress is bigger than the ultimate compressive strength of bone. This is easily done: F 3.6×10 4 N N comp = = =108 2 −4 2 A 3.6 x 10 m m 6 which is less than 170×10
N m
2
. Thus the bone will not break. To find how much shortens we need to use the
definition of the Young's modulus: Y =
F/A l/l o
. Solving for l we find:
1/2
l=
F l0 AY
=
3.6×10 4 N 0.20 m 9
15×10
N m2
3.6×10 m −4
=0.0013 m=1.3 mm 2
Problem II This problem is about a pendulum, which we identify as consisting of Jane and the vine. The period of the pendulum is the time for a full swing back and forth, so the answer we are after is half the period. Then: 1 25 m L T= 2 = =5.0 s 2 2 2 g 9.8 m/ s 1
Problem III Note that to obtain the correct values of the trigonometric functions you need to work with the angles given in radians. (a) The position at any time is x=5.00 cmcos 2t 6 . At t=0, the position of the piston is given by x t=0 =5.00 cmcos =4.33cm , i.e we just evaluated the function in the particular time required by the 6 problem. (b) The velocity is given by the derivative of the position with respect to time: vt =−5.00 cm2Hzsin2t 6 . Thus the velocity at time t=0, is equal to: v t=0 =−5.00 cm2Hzsin 6 =−8.66 cm/ s . 2 (c) Similarly for acceleration we have a t=− 5.00 cm2Hz cos 2t 6 and therefore a t=0 =−5.00 cm2Hz2 cos =−17.32 cm/ s2 . 6 (d) The amplitude is given by the coefficient of the trigonometric function in x(t). Then A=5.00 cm. The period is 2 2 =s . (Remember that the angular frequency is given by the factor that multiplies the just T = = 2 Hz time t in the argument of the trigonometric function)
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3.6 cm2 , (a) will the bone break, and (b) if not, by how much does it shorten? 6 ( The ultimate compressive strength of bone is about 170×10 9 order of 15×10
N m
2
N m
2
and the Young modules of bone is of the
)
Problem II: Jane to the rescue Tarzan stands on a branch as a leopard threatens. Fortunately, Jane is on a nearby branch of the same hight holding a 25mlong vine attached directly above the point midway between her and Tarzan. How soon does she reach Tarzan. Problem III: Piston motion In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x=5.00cm cos2t where x is in centimeters and t is in seconds. At t=0, (a) find the position of the 6 piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion. SOLUTIONS Problem I To know if the bone is going to break we need to find if the stress is bigger than the ultimate compressive strength of bone. This is easily done: F 3.6×10 4 N N comp = = =108 2 −4 2 A 3.6 x 10 m m 6 which is less than 170×10
N m
2
. Thus the bone will not break. To find how much shortens we need to use the
definition of the Young's modulus: Y =
F/A l/l o
. Solving for l we find:
1/2
l=
F l0 AY
=
3.6×10 4 N 0.20 m 9
15×10
N m2
3.6×10 m −4
=0.0013 m=1.3 mm 2
Problem II This problem is about a pendulum, which we identify as consisting of Jane and the vine. The period of the pendulum is the time for a full swing back and forth, so the answer we are after is half the period. Then: 1 25 m L T= 2 = =5.0 s 2 2 2 g 9.8 m/ s 1
Problem III Note that to obtain the correct values of the trigonometric functions you need to work with the angles given in radians. (a) The position at any time is x=5.00 cmcos 2t 6 . At t=0, the position of the piston is given by x t=0 =5.00 cmcos =4.33cm , i.e we just evaluated the function in the particular time required by the 6 problem. (b) The velocity is given by the derivative of the position with respect to time: vt =−5.00 cm2Hzsin2t 6 . Thus the velocity at time t=0, is equal to: v t=0 =−5.00 cm2Hzsin 6 =−8.66 cm/ s . 2 (c) Similarly for acceleration we have a t=− 5.00 cm2Hz cos 2t 6 and therefore a t=0 =−5.00 cm2Hz2 cos =−17.32 cm/ s2 . 6 (d) The amplitude is given by the coefficient of the trigonometric function in x(t). Then A=5.00 cm. The period is 2 2 =s . (Remember that the angular frequency is given by the factor that multiplies the just T = = 2 Hz time t in the argument of the trigonometric function)
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