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Question: A hollow cylinder of radius 4R is rotating about fixed horizontal axis passing through point 'O' wit...

A hollow cylinder of radius 4R is rotating about fixed horizontal axis passing through point 'O' with angular velocity ω0\omega_0. A solid cylinder of radius 'R' is rolling without slipping with respect to inner surface of hollow cylinder. At the given instant, the line OC has angular velocity of 2ω02\omega_0. Point A and B are topmost and bottom most point of solid cylinder respectively and C is its centre. Then, at the given instant

A

Angular velocity of solid cylinder about its own axis is 10ω010 \omega_0

B

Magnitude of acceleration of point A of solid cylinder is 88ω02R88 \omega_0^2 R

C

Magnitude of acceleration of point B of solid cylinder with respect to contact surface of hollow cylinder is 112ω02R112 \omega_0^2 R

D

Radius of curvature of point A is 32R11\frac{32R}{11}

Answer

Magnitude of acceleration of point A of solid cylinder is 88ω02R88 \omega_0^2 R

Explanation

Solution

Let Rh=4RR_h = 4R be the radius of the hollow cylinder and Rs=RR_s = R be the radius of the solid cylinder. The hollow cylinder rotates about O with angular velocity ω⃗h=−ω0k^\vec{\omega}_h = -\omega_0 \hat{k} (clockwise). The line OC rotates about O with angular velocity ω⃗OC=−2ω0k^\vec{\omega}_{OC} = -2\omega_0 \hat{k} (clockwise). The distance OC is Rh−Rs=4R−R=3RR_h - R_s = 4R - R = 3R. At the given instant, C is at (0,−3R)(0, -3R). The velocity of the center of the solid cylinder C is v⃗C=ω⃗OC×r⃗C=(−2ω0k^)×(−3Rj^)=−6ω0Ri^\vec{v}_C = \vec{\omega}_{OC} \times \vec{r}_C = (-2\omega_0 \hat{k}) \times (-3R \hat{j}) = -6\omega_0 R \hat{i}.

Let ω⃗s\vec{\omega}_s be the angular velocity of the solid cylinder about its own axis C. The point of contact P is the bottommost point of the solid cylinder, so r⃗CP=−Rj^\vec{r}_{CP} = -R \hat{j}. The velocity of the point P on the solid cylinder is v⃗P=v⃗C+ω⃗s×r⃗CP\vec{v}_P = \vec{v}_C + \vec{\omega}_s \times \vec{r}_{CP}. The point of contact P on the hollow cylinder is on the inner surface at a distance 4R4R from O, so r⃗OP=−4Rj^\vec{r}_{OP} = -4R \hat{j}. The velocity of the point P on the hollow cylinder is v⃗P′=ω⃗h×r⃗OP=(−ω0k^)×(−4Rj^)=−4ω0Ri^\vec{v}_{P'} = \vec{\omega}_h \times \vec{r}_{OP} = (-\omega_0 \hat{k}) \times (-4R \hat{j}) = -4\omega_0 R \hat{i}. Since there is no slipping, v⃗P=v⃗P′\vec{v}_P = \vec{v}_{P'}. v⃗C+ω⃗s×r⃗CP=v⃗P′\vec{v}_C + \vec{\omega}_s \times \vec{r}_{CP} = \vec{v}_{P'}. −6ω0Ri^+ω⃗s×(−Rj^)=−4ω0Ri^-6\omega_0 R \hat{i} + \vec{\omega}_s \times (-R \hat{j}) = -4\omega_0 R \hat{i}. ω⃗s×(−Rj^)=2ω0Ri^\vec{\omega}_s \times (-R \hat{j}) = 2\omega_0 R \hat{i}. Let ω⃗s=ωsk^\vec{\omega}_s = \omega_s \hat{k}. Then (ωsk^)×(−Rj^)=Rωsi^(\omega_s \hat{k}) \times (-R \hat{j}) = R \omega_s \hat{i}. Rωsi^=2ω0Ri^R \omega_s \hat{i} = 2\omega_0 R \hat{i}, so ωs=2ω0\omega_s = 2\omega_0. The angular velocity of the solid cylinder about its own axis is ω⃗s=2ω0k^\vec{\omega}_s = 2\omega_0 \hat{k} (counter-clockwise). Option (A) is incorrect.

Let's find the acceleration of C. v⃗C=−6ω0Ri^\vec{v}_C = -6\omega_0 R \hat{i}. a⃗C=dv⃗Cdt\vec{a}_C = \frac{d\vec{v}_C}{dt}. The velocity v⃗C\vec{v}_C is given by the rotation of r⃗C\vec{r}_C about O with angular velocity ω⃗OC\vec{\omega}_{OC}. a⃗C=α⃗OC×r⃗C+ω⃗OC×(ω⃗OC×r⃗C)\vec{a}_C = \vec{\alpha}_{OC} \times \vec{r}_C + \vec{\omega}_{OC} \times (\vec{\omega}_{OC} \times \vec{r}_C). Assuming ωOC=2ω0\omega_{OC} = 2\omega_0 is constant, α⃗OC=0\vec{\alpha}_{OC} = 0. a⃗C=ω⃗OC×v⃗C=(−2ω0k^)×(−6ω0Ri^)=12ω02R(k^×i^)=12ω02Rj^\vec{a}_C = \vec{\omega}_{OC} \times \vec{v}_C = (-2\omega_0 \hat{k}) \times (-6\omega_0 R \hat{i}) = 12\omega_0^2 R (\hat{k} \times \hat{i}) = 12\omega_0^2 R \hat{j}.

Let's find the angular acceleration of the solid cylinder. The no-slip condition is v⃗C+ω⃗s×r⃗CP=v⃗P′\vec{v}_C + \vec{\omega}_s \times \vec{r}_{CP} = \vec{v}_{P'}. Differentiating with respect to time: a⃗C+α⃗s×r⃗CP+ω⃗s×v⃗P/C=a⃗P′\vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times \vec{v}_{P/C} = \vec{a}_{P'}. v⃗P/C=ω⃗s×r⃗CP=(2ω0k^)×(−Rj^)=2ω0Ri^\vec{v}_{P/C} = \vec{\omega}_s \times \vec{r}_{CP} = (2\omega_0 \hat{k}) \times (-R \hat{j}) = 2\omega_0 R \hat{i}. a⃗P′\vec{a}_{P'} is the acceleration of the point on the hollow cylinder at P. v⃗P′=−4ω0Ri^\vec{v}_{P'} = -4\omega_0 R \hat{i}. Since ωh=ω0\omega_h = \omega_0 is constant, α⃗h=0\vec{\alpha}_h = 0. a⃗P′=α⃗h×r⃗OP+ω⃗h×(ω⃗h×r⃗OP)\vec{a}_{P'} = \vec{\alpha}_h \times \vec{r}_{OP} + \vec{\omega}_h \times (\vec{\omega}_h \times \vec{r}_{OP}). a⃗P′=0+(−ω0k^)×((−ω0k^)×(−4Rj^))=(−ω0k^)×(−4ω0R(k^×j^))=(−ω0k^)×(4ω0Ri^)=−4ω02R(k^×i^)=−4ω02Rj^\vec{a}_{P'} = 0 + (-\omega_0 \hat{k}) \times ((-\omega_0 \hat{k}) \times (-4R \hat{j})) = (-\omega_0 \hat{k}) \times (-4\omega_0 R (\hat{k} \times \hat{j})) = (-\omega_0 \hat{k}) \times (4\omega_0 R \hat{i}) = -4\omega_0^2 R (\hat{k} \times \hat{i}) = -4\omega_0^2 R \hat{j}.

Substituting into the differentiated no-slip condition: 12ω02Rj^+α⃗s×(−Rj^)+(2ω0k^)×(2ω0Ri^)=−4ω02Rj^12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + (2\omega_0 \hat{k}) \times (2\omega_0 R \hat{i}) = -4\omega_0^2 R \hat{j}. 12ω02Rj^+α⃗s×(−Rj^)+4ω02R(k^×i^)=−4ω02Rj^12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + 4\omega_0^2 R (\hat{k} \times \hat{i}) = -4\omega_0^2 R \hat{j}. 12ω02Rj^+α⃗s×(−Rj^)+4ω02Rj^=−4ω02Rj^12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + 4\omega_0^2 R \hat{j} = -4\omega_0^2 R \hat{j}. 16ω02Rj^+α⃗s×(−Rj^)=−4ω02Rj^16\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) = -4\omega_0^2 R \hat{j}. α⃗s×(−Rj^)=−20ω02Rj^\vec{\alpha}_s \times (-R \hat{j}) = -20\omega_0^2 R \hat{j}. Let α⃗s=αsk^\vec{\alpha}_s = \alpha_s \hat{k}. Then (αsk^)×(−Rj^)=−Rαs(k^×j^)=Rαsi^(\alpha_s \hat{k}) \times (-R \hat{j}) = -R \alpha_s (\hat{k} \times \hat{j}) = R \alpha_s \hat{i}. Rαsi^=−20ω02Rj^R \alpha_s \hat{i} = -20\omega_0^2 R \hat{j}. This equation implies αs=0\alpha_s = 0 and 0=−20ω02Rj^0 = -20\omega_0^2 R \hat{j}, which is a contradiction unless ω0=0\omega_0 = 0. There must be a mistake in setting up the equations or assumptions.

Let's recheck the calculation of a⃗C\vec{a}_C. C is moving in a circle of radius 3R3R about O with angular velocity 2ω02\omega_0. At the instant shown, C is at (0,−3R)(0, -3R), so r⃗C=−3Rj^\vec{r}_C = -3R \hat{j}. The velocity is v⃗C=−6ω0Ri^\vec{v}_C = -6\omega_0 R \hat{i}. The acceleration is purely centripetal towards O, since the speed is constant if ωOC\omega_{OC} is constant. a⃗C=−ωOC2r⃗C=−(2ω0)2(−3Rj^)=4ω02(3Rj^)=12ω02Rj^\vec{a}_C = -\omega_{OC}^2 \vec{r}_C = -(2\omega_0)^2 (-3R \hat{j}) = 4\omega_0^2 (3R \hat{j}) = 12\omega_0^2 R \hat{j}. This matches the previous calculation.

Let's recheck the calculation of a⃗P′\vec{a}_{P'}. P' is moving in a circle of radius 4R4R about O with angular velocity ω0\omega_0. At the instant shown, P' is at (0,−4R)(0, -4R), so r⃗OP′=−4Rj^\vec{r}_{OP'} = -4R \hat{j}. The velocity is v⃗P′=−4ω0Ri^\vec{v}_{P'} = -4\omega_0 R \hat{i}. The acceleration is purely centripetal towards O. a⃗P′=−ωh2r⃗OP′=−(ω0)2(−4Rj^)=4ω02Rj^\vec{a}_{P'} = -\omega_h^2 \vec{r}_{OP'} = -(\omega_0)^2 (-4R \hat{j}) = 4\omega_0^2 R \hat{j}. The calculation of a⃗P′\vec{a}_{P'} was incorrect.

Substituting into the differentiated no-slip condition: a⃗C+α⃗s×r⃗CP+ω⃗s×v⃗P/C=a⃗P′\vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times \vec{v}_{P/C} = \vec{a}_{P'}. 12ω02Rj^+α⃗s×(−Rj^)+(2ω0k^)×(2ω0Ri^)=4ω02Rj^12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + (2\omega_0 \hat{k}) \times (2\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. 12ω02Rj^+α⃗s×(−Rj^)+4ω02Rj^=4ω02Rj^12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + 4\omega_0^2 R \hat{j} = 4\omega_0^2 R \hat{j}. 16ω02Rj^+α⃗s×(−Rj^)=4ω02Rj^16\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) = 4\omega_0^2 R \hat{j}. α⃗s×(−Rj^)=−12ω02Rj^\vec{\alpha}_s \times (-R \hat{j}) = -12\omega_0^2 R \hat{j}. Let α⃗s=αsk^\vec{\alpha}_s = \alpha_s \hat{k}. Then Rαsi^=−12ω02Rj^R \alpha_s \hat{i} = -12\omega_0^2 R \hat{j}. This still leads to a contradiction.

Let's consider the acceleration of P on the solid cylinder relative to C: a⃗P/C=α⃗s×r⃗CP+ω⃗s×(ω⃗s×r⃗CP)\vec{a}_{P/C} = \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}). ω⃗s×r⃗CP=2ω0Ri^\vec{\omega}_s \times \vec{r}_{CP} = 2\omega_0 R \hat{i}. ω⃗s×(ω⃗s×r⃗CP)=(2ω0k^)×(2ω0Ri^)=4ω02Rj^\vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}) = (2\omega_0 \hat{k}) \times (2\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. So, a⃗P=a⃗C+α⃗s×r⃗CP+ω⃗s×(ω⃗s×r⃗CP)\vec{a}_P = \vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}). a⃗P=12ω02Rj^+α⃗s×(−Rj^)+4ω02Rj^=16ω02Rj^+α⃗s×(−Rj^)\vec{a}_P = 12\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) + 4\omega_0^2 R \hat{j} = 16\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}). No slip condition a⃗P=a⃗P′\vec{a}_P = \vec{a}_{P'}. 16ω02Rj^+α⃗s×(−Rj^)=4ω02Rj^16\omega_0^2 R \hat{j} + \vec{\alpha}_s \times (-R \hat{j}) = 4\omega_0^2 R \hat{j}. α⃗s×(−Rj^)=−12ω02Rj^\vec{\alpha}_s \times (-R \hat{j}) = -12\omega_0^2 R \hat{j}. Let α⃗s=αsk^\vec{\alpha}_s = \alpha_s \hat{k}. Then Rαsi^=−12ω02Rj^R \alpha_s \hat{i} = -12\omega_0^2 R \hat{j}. Still a contradiction.

Let's assume the figure is in the vertical plane, and the rotation is about a horizontal axis perpendicular to the plane. Let the x-axis be horizontal to the right, and the y-axis be vertically upwards. Let O be the origin. At the given instant, C is at (0,−3R)(0, -3R). The hollow cylinder rotates clockwise with ω0\omega_0 about O. The line OC rotates clockwise with 2ω02\omega_0 about O. ω⃗h=−ω0k^\vec{\omega}_h = -\omega_0 \hat{k}. ω⃗OC=−2ω0k^\vec{\omega}_{OC} = -2\omega_0 \hat{k}. r⃗C=−3Rj^\vec{r}_C = -3R \hat{j}. v⃗C=ω⃗OC×r⃗C=(−2ω0k^)×(−3Rj^)=−6ω0Ri^\vec{v}_C = \vec{\omega}_{OC} \times \vec{r}_C = (-2\omega_0 \hat{k}) \times (-3R \hat{j}) = -6\omega_0 R \hat{i}. a⃗C=ω⃗OC×v⃗C=(−2ω0k^)×(−6ω0Ri^)=12ω02Rj^\vec{a}_C = \vec{\omega}_{OC} \times \vec{v}_C = (-2\omega_0 \hat{k}) \times (-6\omega_0 R \hat{i}) = 12\omega_0^2 R \hat{j}. Point P on hollow cylinder: r⃗OP=−4Rj^\vec{r}_{OP} = -4R \hat{j}. v⃗P′=ω⃗h×r⃗OP=(−ω0k^)×(−4Rj^)=−4ω0Ri^\vec{v}_{P'} = \vec{\omega}_h \times \vec{r}_{OP} = (-\omega_0 \hat{k}) \times (-4R \hat{j}) = -4\omega_0 R \hat{i}. a⃗P′=ω⃗h×v⃗P′=(−ω0k^)×(−4ω0Ri^)=4ω02Rj^\vec{a}_{P'} = \vec{\omega}_h \times \vec{v}_{P'} = (-\omega_0 \hat{k}) \times (-4\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. Let ω⃗s\vec{\omega}_s be the angular velocity of the solid cylinder about C. Let ω⃗s=ωsk^\vec{\omega}_s = \omega_s \hat{k}. Velocity of P on solid cylinder: v⃗P=v⃗C+ω⃗s×r⃗CP=−6ω0Ri^+(ωsk^)×(−Rj^)=−6ω0Ri^+Rωsi^\vec{v}_P = \vec{v}_C + \vec{\omega}_s \times \vec{r}_{CP} = -6\omega_0 R \hat{i} + (\omega_s \hat{k}) \times (-R \hat{j}) = -6\omega_0 R \hat{i} + R \omega_s \hat{i}. No slip: v⃗P=v⃗P′\vec{v}_P = \vec{v}_{P'}. −6ω0Ri^+Rωsi^=−4ω0Ri^-6\omega_0 R \hat{i} + R \omega_s \hat{i} = -4\omega_0 R \hat{i}. Rωs=2ω0RR \omega_s = 2\omega_0 R, so ωs=2ω0\omega_s = 2\omega_0. ω⃗s=2ω0k^\vec{\omega}_s = 2\omega_0 \hat{k}.

Let α⃗s=αsk^\vec{\alpha}_s = \alpha_s \hat{k} be the angular acceleration of the solid cylinder about C. Acceleration of P on solid cylinder: a⃗P=a⃗C+α⃗s×r⃗CP+ω⃗s×(ω⃗s×r⃗CP)\vec{a}_P = \vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}). a⃗P=12ω02Rj^+(αsk^)×(−Rj^)+(2ω0k^)×((2ω0k^)×(−Rj^))\vec{a}_P = 12\omega_0^2 R \hat{j} + (\alpha_s \hat{k}) \times (-R \hat{j}) + (2\omega_0 \hat{k}) \times ((2\omega_0 \hat{k}) \times (-R \hat{j})). (αsk^)×(−Rj^)=Rαsi^(\alpha_s \hat{k}) \times (-R \hat{j}) = R \alpha_s \hat{i}. (2ω0k^)×((2ω0k^)×(−Rj^))=(2ω0k^)×(2ω0Ri^)=4ω02Rj^(2\omega_0 \hat{k}) \times ((2\omega_0 \hat{k}) \times (-R \hat{j})) = (2\omega_0 \hat{k}) \times (2\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. a⃗P=12ω02Rj^+Rαsi^+4ω02Rj^=Rαsi^+16ω02Rj^\vec{a}_P = 12\omega_0^2 R \hat{j} + R \alpha_s \hat{i} + 4\omega_0^2 R \hat{j} = R \alpha_s \hat{i} + 16\omega_0^2 R \hat{j}. No slip: a⃗P=a⃗P′\vec{a}_P = \vec{a}_{P'}. Rαsi^+16ω02Rj^=4ω02Rj^R \alpha_s \hat{i} + 16\omega_0^2 R \hat{j} = 4\omega_0^2 R \hat{j}. Rαsi^=−12ω02Rj^R \alpha_s \hat{i} = -12\omega_0^2 R \hat{j}. This still gives a contradiction.

Let's check if the direction of ω⃗s\vec{\omega}_s is correct. If the hollow cylinder moves left at the bottom, the solid cylinder should rotate clockwise relative to its center to roll left. So ω⃗s=−ωsk^\vec{\omega}_s = -\omega_s \hat{k}. Then v⃗P=−6ω0Ri^+(−ωsk^)×(−Rj^)=−6ω0Ri^−Rωsi^\vec{v}_P = -6\omega_0 R \hat{i} + (-\omega_s \hat{k}) \times (-R \hat{j}) = -6\omega_0 R \hat{i} - R \omega_s \hat{i}. No slip: −6ω0Ri^−Rωsi^=−4ω0Ri^-6\omega_0 R \hat{i} - R \omega_s \hat{i} = -4\omega_0 R \hat{i}. −Rωs=2ω0R-R \omega_s = 2\omega_0 R, so ωs=−2ω0\omega_s = -2\omega_0. This means ωs\omega_s is 2ω02\omega_0 in the direction of −k^-\hat{k}. So ω⃗s=−2ω0k^\vec{\omega}_s = -2\omega_0 \hat{k}.

Let's recalculate acceleration with ω⃗s=−2ω0k^\vec{\omega}_s = -2\omega_0 \hat{k}. ω⃗s×r⃗CP=(−2ω0k^)×(−Rj^)=−2ω0Ri^\vec{\omega}_s \times \vec{r}_{CP} = (-2\omega_0 \hat{k}) \times (-R \hat{j}) = -2\omega_0 R \hat{i}. ω⃗s×(ω⃗s×r⃗CP)=(−2ω0k^)×(−2ω0Ri^)=4ω02Rj^\vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}) = (-2\omega_0 \hat{k}) \times (-2\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. a⃗P=a⃗C+α⃗s×r⃗CP+ω⃗s×(ω⃗s×r⃗CP)\vec{a}_P = \vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CP} + \vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CP}). a⃗P=12ω02Rj^+(αsk^)×(−Rj^)+4ω02Rj^=Rαsi^+16ω02Rj^\vec{a}_P = 12\omega_0^2 R \hat{j} + (\alpha_s \hat{k}) \times (-R \hat{j}) + 4\omega_0^2 R \hat{j} = R \alpha_s \hat{i} + 16\omega_0^2 R \hat{j}. No slip: Rαsi^+16ω02Rj^=4ω02Rj^R \alpha_s \hat{i} + 16\omega_0^2 R \hat{j} = 4\omega_0^2 R \hat{j}. Rαsi^=−12ω02Rj^R \alpha_s \hat{i} = -12\omega_0^2 R \hat{j}. Still a contradiction.

Let's check option (B). Magnitude of acceleration of point A of solid cylinder is 88ω02R88 \omega_0^2 R. Point A is the topmost point of the solid cylinder. r⃗CA=Rj^\vec{r}_{CA} = R \hat{j}. v⃗A=v⃗C+ω⃗s×r⃗CA=−6ω0Ri^+(−2ω0k^)×(Rj^)=−6ω0Ri^−2ω0R(k^×j^)=−6ω0Ri^−2ω0R(−i^)=−4ω0Ri^\vec{v}_A = \vec{v}_C + \vec{\omega}_s \times \vec{r}_{CA} = -6\omega_0 R \hat{i} + (-2\omega_0 \hat{k}) \times (R \hat{j}) = -6\omega_0 R \hat{i} - 2\omega_0 R (\hat{k} \times \hat{j}) = -6\omega_0 R \hat{i} - 2\omega_0 R (-\hat{i}) = -4\omega_0 R \hat{i}.

a⃗A=a⃗C+α⃗s×r⃗CA+ω⃗s×(ω⃗s×r⃗CA)\vec{a}_A = \vec{a}_C + \vec{\alpha}_s \times \vec{r}_{CA} + \vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CA}). Assuming αs=0\alpha_s = 0. ω⃗s×r⃗CA=(−2ω0k^)×(Rj^)=−2ω0Ri^\vec{\omega}_s \times \vec{r}_{CA} = (-2\omega_0 \hat{k}) \times (R \hat{j}) = -2\omega_0 R \hat{i}. ω⃗s×(ω⃗s×r⃗CA)=(−2ω0k^)×(−2ω0Ri^)=4ω02Rj^\vec{\omega}_s \times (\vec{\omega}_s \times \vec{r}_{CA}) = (-2\omega_0 \hat{k}) \times (-2\omega_0 R \hat{i}) = 4\omega_0^2 R \hat{j}. a⃗A=12ω02Rj^+0+4ω02Rj^=16ω02Rj^\vec{a}_A = 12\omega_0^2 R \hat{j} + 0 + 4\omega_0^2 R \hat{j} = 16\omega_0^2 R \hat{j}. Magnitude is 16ω02R16\omega_0^2 R. This does not match 88ω02R88 \omega_0^2 R.

Let's recheck the calculation of ωs\omega_s. Velocity of P on solid cylinder relative to C: v⃗P/C=ω⃗s×r⃗CP\vec{v}_{P/C} = \vec{\omega}_s \times \vec{r}_{CP}. Velocity of P on hollow cylinder relative to C: v⃗P′/C=v⃗P′−v⃗C=−4ω0Ri^−(−6ω0Ri^)=2ω0Ri^\vec{v}_{P'/C} = \vec{v}_{P'} - \vec{v}_C = -4\omega_0 R \hat{i} - (-6\omega_0 R \hat{i}) = 2\omega_0 R \hat{i}. No slip: v⃗P/C=v⃗P′/C\vec{v}_{P/C} = \vec{v}_{P'/C}. ω⃗s×(−Rj^)=2ω0Ri^\vec{\omega}_s \times (-R \hat{j}) = 2\omega_0 R \hat{i}. Let ω⃗s=ωsk^\vec{\omega}_s = \omega_s \hat{k}. Then Rωsi^=2ω0Ri^R \omega_s \hat{i} = 2\omega_0 R \hat{i}, so ωs=2ω0\omega_s = 2\omega_0. So angular velocity of solid cylinder about its own axis is 2ω02\omega_0. Option (A) is incorrect.

Let's assume there is a mistake in the problem statement or options. Let's review the options and see if any can be derived under some plausible assumptions.

Let's assume the angular acceleration of the solid cylinder is zero, α⃗s=0\vec{\alpha}_s = 0. Then a⃗A=16ω02Rj^\vec{a}_A = 16\omega_0^2 R \hat{j}. ∣a⃗A∣=16ω02R|\vec{a}_A| = 16\omega_0^2 R.

Given that a solution B is provided, and our calculations consistently lead to contradictions or different values, it is highly likely there is an error in the problem statement or the options. However, if forced to choose from the options based on some hidden assumption or interpretation, it's difficult without further information or clarification.

Since I cannot rigorously derive the solution from the given information, I cannot provide a correct and detailed explanation. However, if we assume that the problem is well-posed and option B is correct, then the magnitude of acceleration of point A is 88ω02R88 \omega_0^2 R.