Recitation 10
Solution
Recitation 10 Solution
Problem 1: The cylinder B rolls on the fixed cylinder
A without slipping. If the connected bar CD is rotating
with a constant angular velocity 𝜔𝐶𝐷 = 5𝑟𝑎𝑑/𝑠,
determine:
𝑃
(a) the angular velocity of cylinder B
(b) the acceleration of point D.
Point C is a fixed point.
Solution:
Using rigid body/relative motion formulas to relate points C and D,
𝑣⃑𝐷 = 𝑣⃑𝑐 + 𝜔
⃑⃑𝐶𝐷 × 𝑟⃑𝐷 = 0 + 5𝑘̂ × 0.4𝑖̂ = 2𝑗̂
𝐶
𝑚
𝑠
Also,
𝒎
⃑⃑𝑫 = 𝒂
⃑⃑𝒄 + 𝜶
⃑⃑⃑𝑪𝑫 × 𝒓
⃑⃑𝑫 − 𝝎𝟐𝑪𝑫 𝒓
⃑⃑𝑫 = 𝟎 + 𝟎 + 𝟓𝟐 𝟎. 𝟒𝒊̂ = −𝟏𝟎𝒊̂ 𝟐
𝒂
𝒔
𝑪
𝑪
Since the disk rolls without slip, the points on the two disks instantaneously in contact have the
same velocity. Since the smaller disk is fixed, this is equal to 0.
Then for the larger disk,
𝑣⃑𝐷 = 𝑣⃑𝑃 + 𝜔
⃑⃑𝑃𝐷 × 𝑟⃑𝐷
𝑃
2𝑗̂ = 0 + 𝜔𝑑𝑖𝑠𝑘 𝑘̂ × (0.3)𝑖̂
⇒ 𝝎𝒅𝒊𝒔𝒌 = 𝟔. 𝟔𝟕
𝒓𝒂𝒅
̂ , 𝒑𝒆𝒓 𝑹𝑯𝑹, 𝒌
̂ 𝒊𝒔 𝑪𝑪𝑾
𝒌
𝒔
Recitation 10
Solution
Problem 2: The pin 𝑃 is moving horizontally with a
speed 𝑣𝑃 = 0.5 m/s in a fixed groove that is a height
ℎ = 0.5 m vertically below point 𝐴.
The point 𝑃 pushes the rod 𝐴𝐵 = 1m. The rod 𝐴𝐵
moves the link 𝐵𝐶 = 0.5 m. The link 𝐵𝐶 makes the
block 𝐶 move in the horizontal groove as shown. This
1
groove is 𝑑 = 0.25 +
meters below 𝐴.
√2
𝑆𝑃
At the instant when 𝜃 = 450 , determine:
(a) The angular velocity 𝜔𝐴𝐵 of the bar 𝐴𝐵.
(b) The angular velocity 𝜔𝐵𝐶 of the bar 𝐵𝐶.
(c) The speed 𝑣𝐶 of the block 𝐶.
Given: 𝑣𝑝 , ℎ, 𝑑, 𝐿𝐴𝐵 , 𝐿𝐵𝐶
𝑆𝐶
Solution – Relative motion analysis
All bolded variables are vectors
𝑚
Since point A is fixed, 𝒗𝑨 = 0 𝑠
𝒗𝑩 = 𝒗𝑨 + 𝝎𝑨𝑩 × 𝒓𝑩 = 𝜔𝐴𝐵 𝑘̂ × (𝐿𝐴𝐵 sin 𝜃 𝑖̂ − 𝐿𝐴𝐵 cos 𝜃 𝑗̂) = 𝐿𝐴𝐵 (𝜔𝐴𝐵 cos 𝜃 𝑖̂ + 𝜔𝐴𝐵 sin 𝜃 𝑗̂)
𝑨
Point 𝑃1 is on the link and 𝑃2 is on the pin.
𝒗𝑷𝟏 = 𝒗𝑨 + 𝝎𝑨𝑩 × 𝒓𝑷𝟏 = 𝜔𝐴𝐵 𝑘̂ × (ℎ tan 𝜃 𝑖̂ − ℎ𝑗̂)
𝑨
𝒗𝑷𝟏 = ℎ𝜔𝐴𝐵 𝑖̂ + ℎ𝜔𝐴𝐵 𝑡𝑎𝑛 𝜃𝑗̂ = 𝑣𝑃1,𝑥 + 𝑣𝑃1,𝑦
|𝒗𝑷𝟏 | = ℎ𝜔𝐴𝐵 √1 + tan2 𝜃 = ℎ𝜔𝐴𝐵 sec 𝜃 = 𝑣𝑃1⊥
Under slipping conditions, |𝒗𝑷𝟏 | = 𝑣𝑃1⊥ = 𝑣𝑃 cos 𝜃 = ℎ𝜔𝐴𝐵 sec 𝜃
𝜔𝐴𝐵 =
𝑣𝑃
0.5
𝑟𝑎𝑑
cos2 𝜃 =
cos 2 45 = 0.5
ℎ
0.5
𝑠
𝒗𝑩 = 𝐿𝐴𝐵 𝜔𝐴𝐵 (cos 𝜃 𝑖̂ + sin 𝜃 𝑗̂) = 0.5(cos 45 𝑖̂ + sin 45 𝑗̂)
𝒗𝑪 = 𝒗𝑩 + 𝝎𝑩𝑪 × 𝒓 𝑪 = 0.5(cos 45 𝑖̂ + sin 45 𝑗̂) + −𝜔𝐵𝐶 𝑘̂ × (−𝐿𝐵𝐶 cos 𝜙 𝑖̂ + 𝐿𝐵𝐶 sin 𝜙 𝑗̂)
𝑩
𝒗𝑪 = (0.5 cos 45 + 𝐿𝐵𝐶 𝜔𝐵𝐶 sin 𝜙)𝑖̂ + (−0.5 sin 45 + 𝐿𝐵𝐶 𝜔𝐵𝐶 cos 𝜙)𝑗̂
𝐿𝐴𝐵 cos 𝜃 + 𝐿𝐵𝐶 sin 𝜙 = 𝑑
At 𝜃 = 45𝑜 ,
Recitation 10
Solution
1 ⋅ cos 45 + 0.5 sin 𝜙 = 0.25 +
sin 𝜙 =
1
√2
0.25
= 0.5 ⇒ 𝜙 = 30𝑜
0.5
𝒗𝑪 = (0.5 cos 45 + 0.5𝜔𝐵𝐶 sin 30)𝑖̂ + (−0.5 sin 45 + 0.5𝜔𝐵𝐶 cos 30)𝑗̂
Since slider block C only moves in 𝑖̂ direction, 𝑗̂ direction is zero
0.5 sin 45 + 0.5𝜔𝐵𝐶 cos 30 = 0
𝜔𝐵𝐶 =
0.5 sin 45
2
𝑟𝑎𝑑
=
= 0.8165
0.5 cos 30 √6
𝑠