Let's **solve the problem step by step using the Instantaneous Center (IC) method** as requested. #### **Given:** - Disk radius, \( r = 200\,\text{mm} = 0.2\,\text{m} \) - \( BG = 160\,\text{mm} = 0.16\,\text{m} \) (B is directly left of G, i.e., B is horizontally left of center) - Angular velocity, \( \omega = 8\,\text{rad/s} \) (CCW) - Find the velocity of point A (\( v_A \)) - Point B is to the left of G (center), \( BG \) is horizontal. Let's assume: - **G**: Center of the disk - **A**: Point of contact with the ground (bottommost point) - **B**: 160 mm to the left of G (on the same horizontal as G) --- ## **Step 1: Find the Location of the Instantaneous Center (IC)** For a disk **rolling without slipping**, the instantaneous center (IC) is at the point of contact with the ground, i.e., at point **A**. - **v\(_A\) = 0** (since A is in contact with the ground and not slipping) But the question asks for the **velocity of A**, so perhaps A is not the contact point, or it is asking for the velocity relative to the ground (which would be zero if A is the contact point). But the answer is \( v_A = 1.9305\,\text{m/s} \leftarrow \) (to the left), so perhaps A is a point on the rim of the disk, not the contact point. Let's clarify: - If B is horizontally left of G by 160 mm, and the disk has a radius of 200 mm, then: - The distance from G to A (assuming A is the bottom point): 200 mm down. - The distance from G to B: 160 mm left. Let's **draw the disk**: ``` ^ | (B) |----* | | | G (center) | * (A) ``` - G = center - A = bottom of disk (contact with ground) - B = 160 mm left of G ## **Step 2: IC Location** For rolling without slipping, **IC is at A**, the contact point. ## **Step 3: Velocity of Any Point (Using IC)** The velocity of any point on the body is: \[ v = \omega \cdot r_{IC} \] where \( r_{IC} \) is the **distance from the IC to the point** (perpendicular to the line from IC to the point). But at the **IC (A)**, the velocity is zero. But the question asks for the velocity of A. Since the answer is **not zero**, perhaps A is a different point, not the contact point. **Let's re-express:** - G = center - B = left of G by 160 mm - A = ??? (possibly the topmost point of disk, or a point somewhere else) But the answer is \( v_A = 1.9305\,\text{m/s} \) to the left. Let's try to derive for point B first, then generalize. --- ### **Step 4: Compute the Velocity of Point B (as Example)** - From IC (A, bottom), the distance to G is \( r = 0.2\,\text{m} \) (vertical up) - From IC (A), the distance to B is: - 160 mm left - 200 mm up So, from A to B: \( \vec{r}_{AB} = -0.16\,\hat{i} + 0.2\,\hat{j} \) (from A) - The magnitude: \( r_{AB} = \sqrt{0.16^2 + 0.2^2} = \sqrt{0.0256 + 0.04} = \sqrt{0.0656} = 0.256\,m \) The direction: angle \( \theta = \tan^{-1}(0.2/0.16) = \tan^{-1}(1.25) \approx 51.34^\circ \) above the negative x-axis (left and up). The velocity of B: \[ v_B = \omega \times r_{AB} = 8 \times 0.256 = 2.048\,\text{m/s} \] Direction: perpendicular to \( \vec{r}_{AB} \), i.e., if \( \vec{r}_{AB} \) is left and up, then the velocity is into the page (out of plane) for CCW rotation. But the magnitude matches. --- ### **Step 5: Try to Find the Distance from IC to Point A** But the answer given is \( 1.9305\,\text{m/s} \leftarrow \), which is less than \( v_B \). Suppose point A is at the top of the disk: - From G, up by 200 mm. - From A (IC, bottom of disk), vertical distance from bottom to top = 400 mm = 0.4 m So, from IC to A (top of disk): \( 0.4\,m \) directly up. \[ v_A = \omega \times r_{IC} = 8 \times 0.4 = 3.2\,\text{m/s} \] But this is more than the answer. --- ### **Step 6: Is A Located at an Intermediate Point?** Given the answer is **1.9305 m/s**, let's work backwards: \[ r_{IA} = \frac{v_A}{\omega} = \frac{1.9305}{8} = 0.2413\,\text{m} = 241.3\,\text{mm} \] So, the point A is **241.3 mm from IC**, i.e., from the bottom point. Given the disk radius is 200 mm, and B is 160 mm left of G, perhaps A is at a point displaced from the bottom by some angle. Let’s find the coordinates of A: - Let A be such that the distance from IC (bottom point) to A is 241.3 mm. The general point on the rim of the disk at an angle \( \theta \) from the bottom (measured CCW) has coordinates: \[ \vec{r}_{IA} = r (\sin\theta\,\hat{i} + (1 - \cos\theta)\hat{j}) \] But the magnitude of this vector should be \( 0.2413\,m \). Alternatively, perhaps A is the point **directly horizontally right** of G, i.e., at the 3 o'clock position. Then from the IC (bottom), the distance is: - From G: right by 200 mm - From A (IC): up by 200 mm (to G), then right by 200 mm (to rim at right) - So, from IC to A: \( \sqrt{0.2^2 + 0.2^2} = \sqrt{0.08} = 0.2828\,m \) - \( v_A = 8 \times 0.2828 = 2.262\,\text{m/s} \) Still not matching. --- ### **Step 7: Let's Check for the Point at 160 mm Left of G (Point B)** - From IC: up by 200 mm, left by 160 mm - Distance: \( \sqrt{0.16^2 + 0.2^2} = 0.256\,m \) - \( v = 8 \times 0.256 = 2.048\,\text{m/s} \) Still not matching. --- ### **Step 8: Is A somewhere else?** Given the answer, and the known distances, let's try to relate the geometry. Suppose A is at a position that makes an angle \( \theta \) from the bottom: From IC (bottom), the distance to any point on the rim is: \[ r_{IA} = 2r \sin\left(\frac{\theta}{2}\right) \] Set \( r = 0.2\,m \): \[ 0.2413 = 0.4 \sin\left(\frac{\theta}{2}\right) \implies \sin\left(\frac{\theta}{2}\right) = \frac{0.2413}{0.4} = 0.60325 \implies \frac{\theta}{2} = \arcsin(0.60325) = 37.14^\circ \implies \theta = 74.28^\circ \] So, A is at about 74.28° from the bottom, measured CCW. --- ### **Step 9: Direction of Velocity** The direction of velocity at point A is **perpendicular** to the line from IC to A. The left arrow (\( \leftarrow \)) in the answer means the velocity points to the left. --- ## **Step 10: Final Answer (Summary

Let's carefully solve the problem using the **Instantaneous Center (IC) method**, as requested. --- ### **Given Data:** - Radius of disk: \( r = 200\, \text{mm} = 0.2\, \text{m} \) - Distance \( BG = 160\, \text{mm} = 0.16\, \text{m} \) - The point \( B \) is directly **to the left** of \( G \). - Angular velocity: \( \omega = 8\, \text{rad/s} \) (counterclockwise) - The disk rolls without slipping. - The goal: **Determine the velocity of point A** (unknown location, but given the answer). --- ### **Step 1: Understand the geometry** - \( G \) is the center of the disk. - \( B \) is located **to the left** of \( G \) by 0.16 m. - Since the disk rolls without slipping, **the point of contact \( A \)** (with the surface) is instantaneously at rest relative to the ground. --- ### **Step 2: Identify the Instantaneous Center (IC)** - For pure rolling without slipping, the **IC is at the point of contact \( A \)**. - At the instant shown, the **contact point \( A \)** is momentarily stationary relative to the ground. --- ### **Step 3: Find the position of point \( A \)** - The point \( A \) is at the bottom of the disk, touching the ground. - The center \( G \) is at height \( r = 0.2\, \text{m} \) above the ground. - Point \( B \) is 0.16 m **to the left** of \( G \). --- ### **Step 4: Calculate the velocity of point \( B \)** Using the **IC method**, the velocity of point \( B \) is: \[ v_B = \omega \times r_{IB} \] where \( r_{IB} \) is the **distance from IC to \( B \)**. --- ### **Step 5: Express \( r_{IB} \)** - Since \( B \) is **to the left** of \( G \) by 0.16 m, and \( G \) is at the center, - The **IC** is at the **point of contact \( A \)**, which is **at the bottom** of the disk, touching the ground. - The **center \( G \)** is at height \( r = 0.2\, \text{m} \), - The contact point \( A \) is on the ground directly below \( G \), - The **IC** is at \( A \), on the ground, **horizontally aligned** with \( G \). --- ### **Step 6: Find the position of \( B \) relative to IC** - \( B \) is 0.16 m **to the left** of \( G \), - \( G \) is at the same height as \( G \), - So, the position of \( B \) relative to \( G \): \[ \vec{r}_{BG} = -0.16\, \hat{i} \] - The position of \( G \) relative to \( A \) is: \[ \vec{r}_{GA} = 0\, \hat{i} \quad \text{(since \( G \) is directly above \( A \))} \] - Therefore, the position of \( B \) relative to \( A \): \[ \vec{r}_{BA} = \vec{r}_{BG} + \vec{r}_{GA} = -0.16\, \hat{i} \] --- ### **Step 7: Determine the IC (point of contact) position** - For pure rolling, the **IC** is at \( A \), which is on the ground directly below \( G \). --- ### **Step 8: Find the velocity of \( B \)** Using the IC method: \[ \vec{v}_B = \vec{\omega} \times \vec{r}_{B/\text{IC}} \] - \( \vec{r}_{B/\text{IC}} = \vec{r}_{B} - \vec{r}_{\text{IC}} \) - Since \( \text{IC} \) is at \( A \) (ground contact point), and \( B \) is 0.16 m to the left of \( G \), - The **distance from IC to \( B \)**: \[ r_{B/IC} = |\vec{r}_{B/IC}| = 0.16\,