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1. Velocity of sliding at the beginning of engagement of the two gears = (ωp + ωg) x ______________
a) path of contact
b) arc of contact
c) arc of approach
d) path of approach

View Answer

Answer: d [Reason:] Velocity of sliding at the beginning of engagement = (ωp + ωg) x path of approach. Similarly, the velocity of sliding at the end of engagement = (ωp + ωg) x Path of recess.

2. Velocity of sliding at the pitch point = (ωp + ωg) x _____________
a) path of contact
b) 0
c) arc of approach
d) path of recess

View Answer

Answer: b [Reason:] At the pitch point, there is no relative motion between the gears. Thus, the velocity of sliding is 0. Velocity of sliding at the pitch point = (ωp + ωg) x 0 = 0.

3. The angular velocity of the pinion is 25 rad/s and that of the gear is 10 rad/s. The path of approach is equal to 15 mm. Find the velocity of sliding at the beginning of contact.
a) 394 mm/s
b) 525 mm/s
c) 134 mm/s
d) 348 mm/s

View Answer

Answer: b [Reason:] ωp= 25 rad/s and ωg= 10 rad/s. Path of approach = 15 mm. Velocity of sliding = (ωp + ωg) x Path of approach = (25 + 10) x 15 = 525 mm/s.

4. The angular velocity of the pinion is 40 rad/s and that of the gear is 12 rad/s. The path of recess is equal to 16 mm. Find the velocity of sliding at the end of the contact.
a) 832 mm/s
b) 213 mm/s
c) 1029 mm/s
d) 720 mm/s

View Answer

Answer: a [Reason:] ωp= 40 rad/s and ωg= 12 rad/s. Path of recess = 16 mm. Velocity of sliding = (ωp + ωg) x Path of recess = (40 + 12) x 16 = 832 mm/s.

5. The angular velocity of the pinion is 100 rpm and that of the gear is 30 rpm. The path of recess is equal to 50 mm and the path of contact = 100 mm. Find the velocity of sliding at the beginning of the contact.
a) 1033.29 mm/s
b) 902.93 mm/s
c) 394.02 mm/s
d) 680.58 mm/s

View Answer

Answer: d [Reason:] Np = 100 rpm. Therefore, ωp= 10.47 rad/s Ng = 30 rpm. Therefore, ωg= 3.14 rad/s Path of approach = Path of contact – Path of recess = 100 mm – 50 mm = 50 mm. Velocity of sliding = (ωp + ωg) x Path of approach = (10.47 + 3.14) x 50 = 680.58 mm/s.

6. The angular velocity of the pinion is 200 rpm and that of the gear is 50 rpm. The maximum addendum radius of the larger wheel is 85 mm and the radius of the gear is equal to 75 mm. The pressure angle between the gears is equal to 30°. Find the velocity of sliding at the beginning of the contact.
a) 1289.129 mm/s
b) 732.324 mm/s
c) 453.681 mm/s
d) 239.238 mm/s

View Answer

Answer: c [Reason:] Np = 200 rpm. Therefore, ωp= 20.944 rad/s Ng = 50 rpm. Therefore, ωg= 5.236 rad/s Ra = 85 mm and R = 75 mm; φ = 30° Path of approach = (Ra2 – R2cos2 φ)0.5 – Rsinφ = 17.33 mm Velocity of sliding = (ωp + ωg) x Path of approach = (20.944 + 5.236) x 17.33 = 453.68 mm/s.

7. The angular velocity of the pinion is 325 rpm and that of the gear is 155 rpm. The number of teeth on the larger wheel is 60 and the module is equal to 7 mm. Addendum is 1.1 times the module. The pressure angle is 20°. Find the velocity of sliding at the beginning of the contact.
a) 1010.858 mm/s
b) 1317.858 mm/s
c) 1217.858 mm/s
d) 1117.858 mm/s

View Answer

Answer: a [Reason:] Np = 325 rpm. Therefore, ωp= 34.034 rad/s Ng = 155 rpm. Therefore, ωg= 16.231 rad/s m = 7 mm, T = 60 mm and R = mT/2 = 210 mm Ra = R + a = 210 + 7.7 = 217.7 mm Path of approach= (Ra2 – R2cos2 φ)0.5 – Rsinφ = 20.11 mm Velocity of sliding = (ωp + ωg) x Path of recess = (34.034 + 16.231) x 20.11 = 1010.858 mm/s.

8. The angular velocity of the pinion is 75 rpm and that of the gear is 25 rpm. The addendum radius is 45 mm for the pinion and the radius of the pinion is 40 mm. The pressure angle between the gears is 25°. Find the velocity of sliding at the end of the contact.
a) 89.92 mm/s
b) 91.01 mm/s
c) 94.29 mm/s
d) 97.77 mm/s

View Answer

Answer: d [Reason:] Np = 75 rpm. Therefore, ωp= 7.854 rad/s Ng = 25 rpm. Therefore, ωg= 2.168 rad/s r = 40 mm, ra = 45 mm, φ = 25° Path of recess = (ra2 – r2cos2 φ)0.5 – rsinφ = 9.755 mm Velocity of sliding = (ωp + ωg) x Path of recess = (7.854 + 2.168) x 9.755 = 97.77 mm/s.

9. The angular velocity of the pinion is 150 rpm and that of the gear is 75 rpm. The number of teeth on the pinion is equal to 20 and the module = 6 mm. The value of addendum is equal to 1 module. The pressure angle between the two gears is 20°. Find the velocity of sliding at the end of the contact.
a) 591.29 mm/s
b) 601.12 mm/s
c) 324.87mm/s
d) 512.02 mm/s

View Answer

Answer: c [Reason:] Np = 150 rpm. Therefore, ωp= 15.708 rad/s Ng = 75 rpm. Therefore, ωg= 7.854 rad/s t = 20 and m = 6 mm. r = mt/2 = 60 mm and ra = r + a = 66 mm; φ = 20° Path of recess = (ra2 – r2cos2 φ)0.5 – rsinφ = 13.79 mm Velocity of sliding = (ωp + ωg) x Path of recess = (15.708 + 7.854) x 13.79 = 324.87 mm/s.

10. What is the formula for rolling velocity?
a) ωp x r
b) ωg x R
c) 0
d) ωg x r

View Answer

Answer: a [Reason:] We know that v = r. ω. The rolling velocity is also represented by the same formula ωp x r; where ωp is the angular velocity of the pinion and r is the radius of the smaller gear. Rolling velocity is also called as pitch line velocity.

11. Two involute gears in a mesh have path of approach = 29.83 mm and path of recess = 21.09 mm. The larger gear has 40 teeth and the pinion has 12 teeth. Find the ratio of the sliding to rolling velocity at the:
i) beginning of contact
ii) end of contact
iii) pitch point
r = 72 mm
a) 0.38, 0.54, 0
b) 0.54, 0.38, 0
c) 0.24, 0.63, 0
d) 0.63, 0.24, 0

View Answer

Answer: b [Reason:] i) Ratio of the sliding to rolling velocity = ((ωp + ωg) x Path of approach)/ (ωpx r) = (ωp+ (12/40)ωp) x 29.83/ωp x 72 = 0.54 ii) Ratio of the sliding to rolling velocity = ((ωp + ωg) x Path of recess)/ ωpx r = (ωp+ (12/40)ωp) x 21.09/(ωp x 72) = 0.38 iii) Ratio of the sliding to rolling velocity = ((ωp + ωg) x 0)/ ωpx r = 0.

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