Class Management i Help Homework(Ch.10) Begin Date: 11/7/2017 12:01:00 AM -- Due
ID: 1787309 • Letter: C
Question
Class Management i Help Homework(Ch.10) Begin Date: 11/7/2017 12:01:00 AM -- Due Date: 11/21/2017 11:59:00 PM End Date: 11/21/2017 11:59:00 PM (10%) Problem 4: Suppose an automobile engine can produce 205 Nm of torque, and assume this car is suspended so that the wheels can turn freely. Each wheel acts like a 17.5 kg disk that has a 0.195 m radius. The tires act like 185-kg rings that have inside radii of 0.17 m and outside radii of 0.32 m. The tread of each tire acts like a 9.5-kg hoop of radius 0.35 m. The 14.5-kg axlie acts like a solid cylinder that has a 1.75-cm radius. The 31.5-kg drive shaft acts like a solid cylinder that has a 3.05-cm radius Calculate the angular acceleration, in radians per square second, produced by the motor if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car. Grade Summary Deductions Potential 0% 100% = Submissions Attempts remaining: (3% per attempt) detailed view sinO tan() COS cotanO asinO acos0 atanO acotan)sinh(0 coshOtanh cotanhO 0 END Degrees Radians DEL CLEAR Submit Hint I give up! Hints: 1 % deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedbackExplanation / Answer
Wheel: 17.5 kg disk of 0.195 m radius.
I = ½mr² (disc or cylinder formula)
. = ½ x 17.5 x 0.195²
. = 0.332 kgm²
Tire: 1.85-kg ring, inside radii = 0.17m; outside radius = 0.32 m
I = ½m(r² + r²) (hollow cylinder or ring formula)
. = ½ x 1.85 x (0.17² + 0.32²)
. = 0.1214 kgm²
Tread: 9.5-kg hoop of radius 0.35 m
I = mr² (hoop/ring or cylindrical shell formula)
. = 9.5 x 0.35²
. = 1.163kgm²
Tire: 14.5-kg ring, inside radii = 0.175m
I = ½m(r² ) (hollow cylinder or ring formula)
. = ½ x 14.5 x (0.175²)
. = 0.222 kgm²
Drive shaft: 31.5-kg solid cylinder r = 3.05-cm = 0.0305m.
I = ½mr² (cylinder formula)
. = 31.5 x 0.0305²
. = 0.0293 kgm²
Total moment of intertia
I = 0.332 + 0.1214 + 1.163+ 0.222+0.0293
. = 1.8677kgm²
= /I
. .= 205/1.8677
. .= 109.76 rad/s²